= Solution
Use an <inclusive spectator model>: one heavy constituent decays through the <Fermi interaction>, and sum the open charged-current channels. In <weak-decay channel counting>, each massless final <lepton> pair gives basic width
$$
\Gamma_0(m)=\frac{G_F^2m^5}{192\pi^3},\qquad
\tau=\frac{\hbar}{C\Gamma_0(m)}.
$$
A <quark> pair contributes three colors multiplied by its squared <Cabibbo-Kobayashi-Maskawa matrix> element. Do not multiply by three just because the initial <quark> has three colors: its lifetime is the width of one initial state, averaged over initial color when appropriate. Define
$$
S_u=|V_{ud}|^2+|V_{us}|^2=0.9893,\qquad
S_c=|V_{cd}|^2+|V_{cs}|^2=0.9893.
$$
For the <tau lepton>, two leptonic channels contain an <electron> or a <muon>; light hadronic channels have $d\bar u$ or $s\bar u$. Therefore
$$
C_\tau=2+3S_u=4.9679,\qquad
\Gamma_\tau=2.0103\times10^{-12}\,\mathrm{GeV},\qquad
\boxed{\tau_\tau\simeq3.27\times10^{-13}\,\mathrm s.}
$$
No charmed <hadron> can be produced: the lightest charmed <meson> exceeds the tau <mass>. <Review of Particle Physics by Particle Data Group (2022)>[D-zero and D-plus listings, physical page 1 of each, masses 1864.84 and 1869.66 MeV]{ref} The supplied charm-quark parameter is not a free asymptotic final-particle <mass>. The leading count neglects <electron> and <muon> <masses> and light hadronic threshold effects.
For the <charm quark>, $c\to sW^{+*}$ and $c\to dW^{+*}$ dominate. The emitted current has two light leptonic channels and the same light hadronic channels as above; a tau cannot be produced:
$$
C_c=S_c(2+3S_u)=4.91474347,\qquad
\Gamma_c=8.5232\times10^{-13}\,\mathrm{GeV},\qquad
\boxed{\tau_c^{\mathrm{spectator}}\simeq7.72\times10^{-13}\,\mathrm s.}
$$
For the <bottom quark>, retain $b\to cW^{-*}$ with the supplied $V_{cb}$ and neglect the much smaller $b\to u$ channel. The emitted current can produce three <lepton> flavors and two up-type <quark> families, with three colors for each family. In particular $b\to c\bar cs$ is open at the given <masses>, although suppressed by <phase space>. Counting each open channel as massless gives
$$
C_b=|V_{cb}|^2[3+3S_u+3S_c]=0.01429728,
$$
$$
\Gamma_b=1.0204\times10^{-12}\,\mathrm{GeV},\qquad
\boxed{\tau_b^{\mathrm{spectator}}\simeq6.45\times10^{-13}\,\mathrm s.}
$$
Setting Cabibbo sums to one recovers the familiar counts $5$, $5$ and $9|V_{cb}|^2$. Despite its larger <mass>, the <bottom quark> is slowed by its small interfamily mixing factor.
These are leading estimates rather than precise lifetimes of free <quarks>. <confinement> makes charm and bottom observable in <hadrons>, whose lifetimes involve spectator interactions, <Quantum chromodynamics> corrections and bound-state effects. <mass> corrections are significant for bottom: even the single-charm channel with massless <leptons> has the <three-body weak-decay mass correction>
$$
f(r)=1-8r+8r^3-r^4-12r^2\ln r,\qquad
r=(m_c/m_b)^2=0.09,\qquad f(r)\simeq0.520.
$$
Tau and double-charm channels have different suppressions, so multiplying the entire width by this one factor is inconsistent. Accurate <hadron> lifetimes require those channel-dependent factors and hadronic dynamics; the exercise supports the explicitly stated leading counts.
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