= Solution
At <neutrino decoupling>, the interaction rate is comparable with the <Hubble parameter>. Equating the supplied rates gives
$$
G_F^2T_D^5=\frac{1.66\sqrt{g_*}T_D^2}{m_P},\qquad
\boxed{T_D=\left(\frac{1.66\sqrt{g_*}}{G_F^2m_P}\right)^{1/3}\simeq1.6\,\mathrm{MeV}.}
$$
Here the <Planck mass> is the unreduced $m_P=G^{-1/2}\simeq1.22\times10^{19}\,\mathrm{GeV}$ in <natural units>, $G_F\simeq10^{-5}\,\mathrm{GeV}^{-2}$, and the <effective number of relativistic energy degrees of freedom> is $g_*=10.75$. The supplied rate gives an order-of-magnitude <neutrino decoupling> estimate; a precise neutron-proton conversion temperature uses its own weak-rate coefficients.
The <chemical equilibrium> condition for $n+\nu_e\leftrightarrow p+e^-$ is $\mu_n+\mu_{\nu_e}=\mu_p+\mu_e$. With negligible lepton <chemical potentials>, $\mu_n=\mu_p$. The <nonrelativistic Maxwell--Boltzmann number density> therefore gives
$$
\frac{n_n}{n_p}=\frac{g_n}{g_p}\left(\frac{m_n}{m_p}\right)^{3/2}e^{-(m_n-m_p)/T}
\simeq\boxed{e^{-Q/T}=\frac{X_n}{X_p}},
$$
because $g_n=g_p=2$ and the small mass difference can be neglected in the prefactor. This is the <neutron-proton chemical equilibrium> relation.
For the <deuterium equilibrium abundance>, $\mu_D=\mu_n+\mu_p$ because photons have zero <chemical potential>. The sign convention in the paper is important: $B=m_D-m_n-m_p<0$, while the positive <nuclear binding energy> is $\Delta_D=-B$. Using spin multiplicities $g_D=3$, $g_n=g_p=2$ gives
$$
\frac{n_D}{n_nn_p}=\frac34\left(\frac{2\pi m_D}{m_nm_pT}\right)^{3/2}e^{-B/T}.
$$
With $n_B=\eta n_\gamma$, $m_n\simeq m_p\simeq m_N$, $m_D\simeq2m_N$ and <photon number density> $n_\gamma=2\zeta(3)T^3/\pi^2$, this becomes
$$
\boxed{X_D\simeq\frac{12\zeta(3)}{\sqrt\pi}\,\eta X_nX_p\left(\frac{T}{m_N}\right)^{3/2}e^{-B/T}.}
$$
Here $X_D=n_D/n_B$ is a number fraction; the fraction of baryons bound in <deuterium> is $2X_D$. The formula estimates the onset of the <deuterium bottleneck> opening while depletion of free <neutrons> and <protons> remains small. Because $\eta\ll1$, stable <deuterium> becomes abundant only far below the positive <nuclear binding energy>, not simply at $T=\Delta_D$.
Let $T_f$ denote <cosmological weak freeze-out> of the <neutron-to-proton ratio>, $r_f=e^{-Q/T_f}$, and let $t_N$ mark efficient nuclear burning. <Free-neutron decay after freeze-out> gives
$$
X_n(t_N)\simeq\frac{r_f}{1+r_f}e^{-(t_N-t_f)/\tau_n},\qquad\tau_n=\frac{t_{1/2}}{\log2}.
$$
Almost every surviving <neutron> enters helium-4, whose <atomic nucleus> contains two <neutrons> and four baryons. Thus <neutron-limited helium synthesis> gives
$$
\boxed{Y_p\simeq2X_n(t_N)=\frac{2r_f}{1+r_f}e^{-(t_N-t_f)/\tau_n}.}
$$
A longer neutron <half-life> leaves more <neutrons> available, increasing the <primordial helium mass fraction>. If the longer lifetime reflects weaker <weak interactions>, the accompanying earlier <cosmological weak freeze-out> can reinforce that effect.
During <radiation domination>, $H\propto\sqrt{Gg_*}\,T^2$ whereas the weak conversion rate is proportional to $T^5$. Hence $T_f\propto(Gg_*)^{1/6}$. Increasing the <effective number of relativistic energy degrees of freedom> raises $T_f$, increasing $r_f$, and shortens the interval for <free-neutron decay after freeze-out>. Both increase the <primordial helium mass fraction> while nuclear burning remains efficient. The same reasoning gives the <primordial helium response to a varying gravitational constant>: larger $G$ over this interval produces more helium, with fixed microscopic rates and the approximate <Friedmann equation> unchanged. Smaller $G$ has the opposite effect. A particular varying-$G$ model must specify its background expansion and time history; the simple trend must not be extrapolated to changes so large that nuclear burning itself fails.
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