= Solution
Use <natural units> $\hbar=c=k_B=1$ and the unreduced <Planck mass> $M_P=G^{-1/2}$. <Neutrino decoupling> occurs when the <weak interaction> rate falls below the <Hubble parameter>. The <weak-decoupling temperature estimate> equates the given rate with the radiation-era expansion rate:
$$
G_F^2T_d^5\simeq1.66\sqrt{g_*}\frac{T_d^2}{M_P},\qquad \boxed{T_d\simeq\left(\frac{1.66\sqrt{g_*}}{G_F^2M_P}\right)^{1/3}\simeq1.8\ {\rm MeV}.}
$$
Here $g_*$ counts the <effective number of relativistic energy degrees of freedom>, and the <Fermi constant> is $G_F\simeq10^{-5}\ {\rm GeV}^{-2}$. The omitted reaction-rate coefficients make this an order-of-magnitude estimate: \b[decoupling occurs at a temperature of order one MeV].
The charge-conserving channel is $n+\nu_e\leftrightarrow p+e^-$, together with $p+\bar\nu_e\leftrightarrow n+e^+$. The positron sign in the printed neutron-capture channel violates <electric charge conservation>. <Chemical equilibrium> requires $\mu_n+\mu_{\nu_e}=\mu_p+\mu_e$. Negligible lepton <chemical potentials> therefore give $\mu_n\simeq\mu_p$. Since <neutrons> and <protons> are <nonrelativistic particles>, their <Maxwell-Boltzmann distribution> gives
$$
n_i=g_i\left(\frac{m_iT}{2\pi}\right)^{3/2}e^{(\mu_i-m_i)/T},\qquad \frac{n_n}{n_p}=\frac{g_n}{g_p}\left(\frac{m_n}{m_p}\right)^{3/2}e^{-(m_n-m_p)/T}.
$$
Both nucleons have two spin states, and their masses agree closely in the prefactor. Thus <neutron-proton chemical equilibrium> implies
$$
\boxed{\frac{n_n}{n_p}=\frac{X_n}{X_p}\simeq e^{-Q/T_d},\qquad Q=m_n-m_p>0.}
$$
The heavier <neutron> is suppressed by a <Boltzmann factor>.
For the <deuterium equilibrium abundance>, retain the paper's convention $B=m_D-m_n-m_p<0$; the positive <nuclear binding energy> is $\varepsilon_D=-B\simeq2.22\ {\rm MeV}$. <Chemical equilibrium> of $n+p\leftrightarrow D+\gamma$ gives $\mu_D=\mu_n+\mu_p$ because the equilibrium <photon> <chemical potential> vanishes. Using the <nonrelativistic Maxwell--Boltzmann number density> for each species and dividing gives
$$
\frac{n_D}{n_nn_p}=\frac{g_D}{g_ng_p}\left(\frac{m_D}{m_nm_p}\right)^{3/2}\left(\frac{2\pi}{T}\right)^{3/2}e^{-B/T}.
$$
Here $g_D=3$ and $g_n=g_p=2$. With the <baryon-to-photon ratio> $\eta=n_B/n_\gamma$, the <photon number density> $n_\gamma=2\zeta(3)T^3/\pi^2$, and $m_D\simeq2m_N$, $m_n\simeq m_p\simeq m_N$, this becomes
$$
\boxed{\frac{X_D}{X_nX_p}\simeq\frac{12\zeta(3)}{\sqrt\pi}\,\eta\left(\frac{T}{m_N}\right)^{3/2}e^{-B/T}\simeq8.14\,\eta\left(\frac{T}{m_N}\right)^{3/2}e^{\varepsilon_D/T}.}
$$
In this convention $X_D=n_D/n_B$ counts nuclei, so its contribution to the fraction of baryons in <deuterium> is $2X_D$.
The <deuterium bottleneck> explains why <helium-4> formation waits until $T$ is far below $\varepsilon_D$. At higher <temperature>, there are vastly more <photons> than baryons; even the dilute energetic tail can destroy newly formed <deuterium>. A crude estimate balances $\eta^{-1}e^{-\varepsilon_D/T}$ against one, giving $T\sim\varepsilon_D/\log(\eta^{-1})\sim0.1\ {\rm MeV}$ for $\eta$ of order $10^{-9}$. The power-law factors in the <deuterium equilibrium abundance> and the nuclear reaction rates refine this estimate. Once stable <deuterium> accumulates, further reactions can efficiently place nearly all surviving <neutrons> into <helium-4>.
Write $r_N=n_n/n_p$ at the onset of nuclear burning. Each <helium-4> nucleus requires two <neutrons> and contains four baryons. <Neutron-limited helium synthesis> therefore gives the <primordial helium mass fraction>
$$
\boxed{Y_4\simeq\frac{4n_{\rm He}}{n_B}\simeq2X_n(T_N)=\frac{2r_N}{1+r_N}.}
$$
For a realistic rough <cosmological weak freeze-out> estimate, $r_f$ is about $1/6$ and <free-neutron decay after freeze-out> reduces it to about $1/7$, giving \b[roughly one quarter of the baryonic mass in helium-4]. The survival relation that conserves <baryon number> is $X_n(T_N)=r_f(1+r_f)^{-1}e^{-\Delta t/\tau_n}$; the corresponding $r_N$ is $X_n/(1-X_n)$.
The rough <neutrino decoupling> rate by itself does not determine that one-quarter value. Substituting $T_d=1.8\ {\rm MeV}$ literally into $e^{-Q/T_d}$ would give $r\simeq0.49$ and $Y_4\simeq0.65$ before neutron decay. <Neutron-proton chemical equilibrium> persists to a lower temperature, approximately $0.7$--$0.8\ {\rm MeV}$ once the actual conversion rates are included. The rate scaling fixes the MeV scale; accurate abundances need the conversion coefficients and the delay through the <deuterium bottleneck>. This sequence is described in https://www.davidtong.org/teaching/cosmology/cosmohtml/S2#S2.SS5.SSS3[David Tong's discussion of nucleosynthesis].
Increasing $Q$ reduces the equilibrium <neutron-to-proton ratio>, so \b[the primordial helium abundance decreases]. Holding the freeze-out temperature $T_f$ and the decay survival factor fixed, the <primordial helium response to the neutron-proton mass difference> is
$$
\boxed{\frac{Y_4(1.1Q)}{Y_4(Q)}\simeq\frac{1+e^{Q/T_f}}{1+e^{1.1Q/T_f}}.}
$$
For $Q\simeq1.29\ {\rm MeV}$ and $T_f=0.7$--$0.8\ {\rm MeV}$ this is about $0.85$--$0.87$, a reduction of roughly fifteen percent. A larger $Q$ also changes neutron decay and detailed weak rates; their quantitative effects require more microscopic information. The sign from the <Boltzmann factor> is already clear.
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