= Solution
\b[The nuclear determination of $|V_{ud}|$]. A $0^+\to0^+$ <beta decay transition> has no nuclear <parity> change and can proceed with zero lepton <orbital angular momentum>. The allowed scalar operator gives a <Fermi transition>; the rank-one <Gamow-Teller transition> operator cannot connect total nuclear spins zero to zero. The strong isobaric-analog transitions used for precision measurements are <superallowed beta decays>. A generic $0^+\to0^+$ assignment alone would not guarantee superallowed strength if the states were nonanalog.
At the <quark> level the <weak charged current> converts $d\leftrightarrow u$ and carries $V_{ud}$ in the amplitude. The nuclear Fermi operator is an <isospin> ladder operator, giving
$$
\mathcal M\propto G_FV_{ud}M_F,\qquad
\boxed{\Gamma\propto G_F^2|V_{ud}|^2|M_F|^2f.}
$$
Here $f$ denotes the known <phase space> dependence, with the dimensional factors included in the proportionality constant. In the isospin-symmetric limit, $M_F$ between analog states is fixed by isospin algebra, for example $|M_F|^2=2$ for the usual $T=1$ analog transitions. A measured partial <half-life> and the computed <phase space> factor give the <comparative half-life>; after <radiative corrections> and <isospin> breaking corrections, its inverse determines $|V_{ud}|^2$. The decay rate measures a magnitude squared, not the complex phase of $V_{ud}$.
\b[Ds decay diagrams and CKM comparisons]. The original PDF has $K^+\bar K^0$, with $\bar K^0=s\bar d$. The local TeX drops that overbar; using $K^0=d\bar s$ instead would give a different flavour channel. Use the PDF's <neutral antikaon> throughout.
In the representative <Ds nonleptonic tree amplitudes>, the <Ds meson> spectator is $\bar s$. The elementary weak transitions are
$$
\begin{array}{c|c|c}
\text{Channel}&\text{Quark transition}&\text{CKM magnitude}\\\hline
K^+\pi^0&c\to dW^+,\quad W^+\to u\bar d&|V_{cd}V_{ud}|\\
K^+\bar K^0&c\to sW^+,\quad W^+\to u\bar d&|V_{cs}V_{ud}|\\
\phi\pi^+&c\to sW^+,\quad W^+\to u\bar d&|V_{cs}V_{ud}|
\end{array}
$$
\Image[/past-exam-of-the-particle-and-nuclear-physics-course-of-the-university-of-cambridge/2018-ds-nonleptonic-diagrams.png]
{title=The three Ds spectator tree diagrams, using the neutral-antikaon channel in the original PDF}
In the first two channels, the produced $u$ joins the spectator $\bar s$ to form $K^+$; the other quark joins $\bar d$. For the first channel this uses the $-d\bar d/\sqrt2$ component of the <neutral pion>. In the third, $s\bar s$ forms the <phi meson> and $u\bar d$ forms $\pi^+$. The two possible final pairings have different colour contractions and hadronic <matrix elements> even when their weak CKM factors coincide.
The purely CKM rate suppression of $c\to d$ relative to $c\to s$ is
$$
\boxed{\left|\frac{V_{cd}}{V_{cs}}\right|^2
=\left(\frac{0.225}{0.974}\right)^2=0.0533638.}
$$
Thus $K^+\pi^0$ is singly <Cabibbo suppressed>, while $K^+\bar K^0$ and $\phi\pi^+$ are Cabibbo-favored at this level. If reduced hadronic strengths and <phase space> effects were equal, including the neutral-pion component would give
$$
\left[\frac{B(K^+\pi^0)}{B(K^+\bar K^0)}\right]_{\rm simple}
=\frac12\left|\frac{V_{cd}}{V_{cs}}\right|^2=0.0266819.
$$
The supplied <branching fractions> instead give
$$
\frac{B(K^+\pi^0)}{B(K^+\bar K^0)}=\frac{6.3\times10^{-4}}{2.9\times10^{-2}}=0.0217241,
\qquad
\frac{B(K^+\pi^0)}{B(\phi\pi^+)}=0.0140,
$$
$$
\frac{B(K^+\bar K^0)}{B(\phi\pi^+)}=0.64444.
$$
The first observed ratio is $0.814$ times the neutral-pion-corrected simple estimate, requiring only an order-one remaining hadronic and <phase space> factor. Relative to $\phi\pi^+$, that remaining factor is $0.525$. The two Cabibbo-favored channels differ by a factor $0.644$ despite equal CKM magnitudes. The CKM hierarchy therefore explains why one channel is much rarer, but it does not calculate the three exclusive <branching fractions>; <spin>, <phase space>, <color charge> and <strong interaction> final-state effects also matter.
\b[What spin and parity establish]. In either pseudoscalar-pseudoscalar final state, the total intrinsic <spin> is zero. Conservation of <angular momentum> for a spin-zero parent forces $L=0$, so the final <parity> is
$$
P_f=(-1)(-1)(-1)^0=+1,
$$
opposite to the parent $0^-$ <Ds meson>. Consequently
$$
\boxed{D_s^+\to K^+\pi^0,\ K^+\bar K^0\text{ require parity violation and are weak decays}.}
$$
For a <vector meson> plus a <pion>, the intrinsic final <spin> is one. Coupling to total $J=0$ requires $L=1$, giving $P_f=(-1)(-1)(-1)^1=-1$, equal to the parent parity. Thus \b[spin and parity alone permit a parity-conserving $\phi\pi^+$ transition]. That is the limited sense in which it is not necessarily weak from this information alone. In reality it is also weak: the initial charm flavour disappears, which the <strong interaction> and <electromagnetic interaction> conserve.
\b[W branching fractions]. In the massless-daughter, tree-level estimate, each leptonic channel contributes one common unit of <decay width>. Each accessible quark pair contributes $3|V_{ij}|^2$ because of the three final colours. The <charged-current W branching-fraction counting> denominator is
$$
D=3+3\sum_{i=u,c}\sum_{j=d,s,b}|V_{ij}|^2
=9.000897\simeq9.
$$
Only $u$ and $c$ are accessible up-type quarks; the large $V_{tb}$ is irrelevant because a <W boson> cannot decay into the heavier <top quark>. Hence
$$
\boxed{B(W^-\to e^-\bar\nu_e)=\frac1D\simeq\frac19=11.1\%.}
$$
A final <bottom quark> arises in $W^-\to b\bar u$ or $b\bar c$, so
$$
\boxed{B(W^-\text{ containing }b)=\frac{3(|V_{ub}|^2+|V_{cb}|^2)}D
=\frac{3(0.004^2+0.041^2)}{9.000897}
=5.66\times10^{-4}\simeq0.057\%.}
$$
This counts the primary decay channels; finite daughter masses and <QCD> corrections refine the approximation.
\b[The requested top-lifetime estimate and its limitation]. Under the instructed equal-coefficient <Sargent's rule> scaling, $\Gamma\propto m^5$ and the <mean lifetime> is inversely proportional to that fifth power. Converting both masses to GeV gives
$$
\boxed{\tau_t^{(m^5)}=\tau_\mu\left(\frac{m_\mu}{m_t}\right)^5
=2.2\times10^{-6}\,\mathrm s\left(\frac{0.106}{175}\right)^5
=1.79\times10^{-22}\,\mathrm s.}
$$
This is the requested order-of-magnitude estimate, assuming the same effective coefficient for the two total decay rates. It corresponds to $c\tau\simeq54\,\mathrm{fm}$ and is longer than the <hadronization time>, roughly a few $10^{-24}\,\mathrm s$. Taken literally, it would allow top-flavored hadrons to form; it does not establish the physical statement that the top decays before hadronizing.
The extrapolation fails because <Sargent's rule> uses a low-energy <Fermi interaction> approximation, whereas $m_t>m_W+m_b$ permits $t\to bW$ with an on-shell <W boson>. The physical leading <decay width> instead scales as
$$
\Gamma(t\to bW)\simeq\frac{G_Fm_t^3|V_{tb}|^2}{8\pi\sqrt2}
(1-x)^2(1+2x),\qquad x=\frac{m_W^2}{m_t^2}.
$$
Using $m_W\simeq80.4\,\mathrm{GeV}$, the <Fermi constant> $G_F=1.16638\times10^{-5}\,\mathrm{GeV^{-2}}$ and $|V_{tb}|=0.999$ gives about $1.55\,\mathrm{GeV}$ at the supplied top mass. Then $\tau_t=\hbar/\Gamma_t$, with $\hbar=6.58212\times10^{-25}\,\mathrm{GeV\,s}$, gives $\tau_t\simeq4.2\times10^{-25}\,\mathrm s$. This corrected regime explains <top-quark decay before hadronization>.<Review of Particle Physics by Particle Data Group (2022)>[Top Quark review, section 61.2, equation 61.1 and comparison of lifetime with hadron formation]{ref}
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