= Solution
Use <natural units> with $c=\hbar=k_B=1$ and the unreduced <Planck mass> $M_{\rm Pl}=G^{-1/2}$. In a <radiation-dominated universe>, the <effective number of relativistic energy degrees of freedom> $g_*=\mathcal N$ gives
$$
\rho_r=\frac{\pi^2}{30}g_*T^4,\qquad
H^2=\frac{8\pi G}{3}\rho_r,\qquad
H=\sqrt{\frac{8\pi^3}{90}}\frac{\sqrt{g_*}T^2}{M_{\rm Pl}}
\simeq1.66\frac{\sqrt{g_*}T^2}{M_{\rm Pl}}.
$$
The ratio of weak conversion rate to expansion rate is proportional to $T^3$, so it decreases as the universe cools. <Cosmological weak freeze-out> occurs approximately at $\Gamma=H$, giving
$$
\boxed{T_d\simeq
\left(\frac{1.66\sqrt{g_*}}{G_F^2M_{\rm Pl}}\right)^{1/3}
\sim1\text{--}2\ {\rm MeV}.}
$$
For $g_*=10.75$, $G_F=10^{-5}\ {\rm GeV}^{-2}$ and $M_{\rm Pl}=1.22\times10^{19}\ {\rm GeV}$ this estimate is $1.65\ {\rm MeV}$. Equivalently $g_*^{1/6}\simeq1.5$ is only a mild multiplicative correction. The rate itself was given only to order of magnitude, so this is a decoupling scale, not a precise prediction for the weak freeze-out abundance.
Both <neutrons> and <protons> are nonrelativistic at this <temperature>. The <nonrelativistic Maxwell--Boltzmann number density> follows by the Gaussian momentum integral:
$$
n_i=g_i\int\frac{d^3p}{(2\pi)^3}
e^{-(m_i+p^2/(2m_i)-\mu_i)/T}
=g_i\left(\frac{m_iT}{2\pi}\right)^{3/2}e^{(\mu_i-m_i)/T}.
$$
Their equal spin degeneracies are $g_n=g_p=2$. <Chemical equilibrium> of the weak reaction gives $\mu_n+\mu_{\nu_e}=\mu_p+\mu_e$; with negligible lepton asymmetry, $\mu_n\simeq\mu_p$. Taking the ratio and setting the nearly equal nucleon masses equal only in the prefactor gives the <neutron-proton chemical equilibrium> result
$$
\frac{n_n}{n_p}
=\left(\frac{m_n}{m_p}\right)^{3/2}
e^{(\mu_n-\mu_p-Q)/T}
\simeq e^{-Q/T},\qquad
\boxed{\frac{X_n}{X_p}\simeq e^{-Q/T_d}.}
$$
The common baryon normalization cancels. Keeping the exact mass prefactor changes this by only order $Q/m_p$. Thus the printed simple exponential is the usual nonrelativistic equal-mass-prefactor approximation. Using $Q\simeq1.29\ {\rm MeV}$ with the rough decoupling scale above gives a ratio of order one half; the rate's omitted numerical factors and gradual freeze-out must be included before using a more accurate neutron abundance.
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