= Solution
During <radiation domination>, the <effective number of relativistic energy degrees of freedom>, denoted $g_*$, gives the <energy density>
$$
\rho_R=\frac{\pi^2}{30}g_*T^4.
$$
The paper's $\mathcal N_{\rm eff}$ is this $g_*$; it is not the conventionally normalized effective number of <neutrino> species. With the specified <reduced Planck mass>, the <Friedmann equation> reads $H^2=\rho_R/(3M_{\rm Pl}^2)$, whence
$$
H=\sqrt{\frac{\pi^2g_*}{90}}\frac{T^2}{M_{\rm Pl}}.
$$
Write the <weak interaction> conversion rate as $\Gamma=A G_F^2T^5$. The ratio $\Gamma/H$ scales as $T^3$, so it falls during cooling: conversion eventually becomes too slow to maintain <neutron-proton chemical equilibrium>. Equating the conversion and expansion rates gives
$$
T_F=\left[\frac{\sqrt{\pi^2g_*/90}}{A G_F^2M_{\rm Pl}}\right]^{1/3}.
$$
On suppressing the numerical coefficient, the <cosmological weak freeze-out> <temperature> is consequently
$$
\boxed{T_F\sim G_F^{-2/3}\mathcal N_{\rm eff}^{1/6}M_{\rm Pl}^{-1/3}.}
$$
The one-sixth power reflects a balance between a conversion rate proportional to $T^5$ and a <Hubble parameter> proportional to $\sqrt{g_*}\,T^2$.
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