= Solution
In both <semileptonic decays>, an active <quark> emits a virtual <W boson>, which produces an <Electron> and an <electron antineutrino>:
$$
\begin{aligned}
n:&\quad d\to u+W^-,\quad W^-\to e^-+\bar\nu_e,\\
\Lambda^0:&\quad s\to u+W^-,\quad W^-\to e^-+\bar\nu_e.
\end{aligned}
$$
The unchanged $u,d$ <spectator quarks> combine with the produced $u$ into the final <proton>. The relevant <Cabibbo-Kobayashi-Maskawa matrix> elements are $V_{ud}\simeq\cos\theta_C$ and $V_{us}\simeq\sin\theta_C$.
\Image[/past-exam-of-the-particle-and-nuclear-physics-course-of-the-university-of-cambridge/2019-beta-cabibbo-diagrams.png]
{title=Neutron and Lambda semileptonic decay diagrams and their Cabibbo factors}
Use the quoted <masses> in <energy> units, equivalently <natural units> $c=1$, and define the energy releases before subtracting the electron rest energy as
$$
\Delta_n=939.6-938.3=1.3\,\mathrm{MeV},\qquad
\Delta_\Lambda=1115.7-938.3=177.4\,\mathrm{MeV}.
$$
The prescribed <density of states> ansatz uses precisely these differences to the fifth power. A <Cabibbo-angle extraction from semileptonic decay rates> also needs a common reduced hadronic strength. In that approximation the partial decay rates, measured in $\mathrm s^{-1}$, are
$$
r_n=K\cos^2\theta_C\,\Delta_n^5,\qquad
r_\Lambda=K\sin^2\theta_C\,\Delta_\Lambda^5.
$$
The supplied <mean lifetimes> describe all decays, so multiply their inverses by the relevant <branching fractions>:
$$
r_n=\frac{1}{880\,\mathrm s}=1.13636\times10^{-3}\,\mathrm s^{-1},\qquad
r_\Lambda=\frac{8.32\times10^{-4}}{2.63\times10^{-10}\,\mathrm s}
=3.16350\times10^6\,\mathrm s^{-1}.
$$
Taking their ratio cancels $K$ and gives
$$
\tan^2\theta_C
=\frac{r_\Lambda}{r_n}\left(\frac{\Delta_n}{\Delta_\Lambda}\right)^5
=\frac{8.32\times10^{-4}\times880}{2.63\times10^{-10}}
\left(\frac{1.3}{177.4}\right)^5
=0.0588300.
$$
Choose the conventional first-quadrant <Cabibbo angle>:
$$
\boxed{\theta_C=\arctan\sqrt{0.0588300}=0.237954\,\mathrm{rad}\simeq13.6^\circ.}
$$
The <Lambda baryon> partial rate is much larger despite its smaller weak flavour coupling because the fifth-power <phase space> factor is enormous.
There is an assumption beyond the stated <density of states> ansatz: the neutron and Lambda reduced hadronic strengths have been set equal. If their ratio is $H=K_\Lambda/K_n$, the same data instead imply $\tan^2\theta_C=0.0588300/H$. Without specifying $H$, the angle is not uniquely determined. Unequal <hadronic form factors>, finite <electron mass> and recoil make the displayed common-strength result an exercise-level estimate, rather than a precision measurement.
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