= Solution
The <effective number of relativistic energy degrees of freedom> count two photon polarizations, four electron/positron spin states, and six neutrino/antineutrino states, with the <Fermi-Dirac distribution> factor $7/8$ for <fermions>. Treating these species as relativistic and at a common temperature gives $\boxed{g_*=2+(7/8)(4+6)=10.75}$.
The mass-equivalent <radiation mass density> is $\rho=(\pi^2/30)g_*(k_BT)^4/(\hbar^3c^5)$. Therefore the <Friedmann equation> gives
$$
H(T)=\left(\frac{4\pi^3Gg_*}{45\hbar^3c^5}\right)^{1/2}(k_BT)^2.
$$
At $T_0=10^{10}\,\mathrm K$, $H_0\simeq0.503\,\mathrm{s}^{-1}$. Equating the specified <weak interaction> rate to $H_0(T/T_0)^2$ gives $(T_*/T_0)^3=H_0/(1\,\mathrm{s}^{-1})$ and hence
$$
\boxed{T_*\simeq7.96\times10^9\,\mathrm K,\qquad T_*\propto g_*^{1/6}.}
$$
The approximate equilibrium <neutron-to-proton ratio>, neglecting chemical potentials, is $r_*\simeq\exp[-(m_n-m_p)c^2/(k_BT_*)]$, about $0.152$ using the mass difference $1.293\,\mathrm{MeV}$. This is a freeze-out estimate rather than a detailed reaction-network computation.
After weak <cosmological weak freeze-out>, free neutrons undergo <beta-minus decay>; neutrinos decouple, and electron-positron annihilation heats the photons relative to the neutrinos. Initially the abundant energetic photons destroy deuterium almost as soon as it forms. Once the <deuterium bottleneck> ends, nuclear reactions rapidly pass through deuterium, tritium and helium-three to <helium-4>. Reactions eventually become too slow compared with <cosmic expansion>; nearly all surviving neutrons are bound in helium-four. Its mass fraction is approximately $Y_4=2r/(1+r)$, where $r$ is evaluated when nuclear assembly begins.
Increasing $g_*$ increases the <Hubble parameter>, raises $T_*$, increases the freeze-out <neutron-to-proton ratio>, and leaves less time for <free-neutron decay after freeze-out>. Both effects increase $Y_4$; decreasing $g_*$ reverses them. Most deuterium is only an intermediate product and is burned into helium, so its residual abundance and its absolute change are tiny compared with the helium mass fraction. \b[This does not imply a negligible fractional change in residual deuterium.] Faster expansion also shortens the time available to burn it, and its relative abundance can be appreciably sensitive to extra radiation. The distinction matters when interpreting the final wording of the question; joint deuterium and helium abundances constrain extra relativistic species, as discussed in https://arxiv.org/abs/1412.1408[the Particle Data Group nucleosynthesis review].
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