= Solution
In the nonrelativistic dilute regime, $E(p)=mc^2+p^2/(2m)+\cdots$ and $(mc^2-\mu)/kT\gg1$, so the Bose or Fermi correction in the denominator is negligible. The common <Maxwell-Boltzmann distribution> approximation gives
$$
n\simeq\frac{4\pi g_s}{h^3}e^{(\mu-mc^2)/kT}\int_0^\infty p^2e^{-p^2/(2mkT)}dp.
$$
Differentiating the quoted <Gaussian integral> with respect to its squared scale gives $\int_0^\infty p^2e^{-ap^2}dp=\sqrt\pi/(4a^{3/2})$. Therefore
$$
\boxed{n=g_s\left(\frac{2\pi mkT}{h^2}\right)^{3/2}e^{(\mu-mc^2)/kT}.}
$$
<chemical equilibrium> of $p+n\leftrightarrow D$ requires $\mu_D=\mu_p+\mu_n$. Apply the density formula to each species to obtain
$$
\frac{n_D}{n_nn_p}=\frac{g_D}{g_ng_p}\left(\frac{2\pi kT}{h^2}\right)^{-3/2}
\left(\frac{m_D}{m_nm_p}\right)^{3/2}e^{B_D/kT}.
$$
Using $g_n=g_p=2$, the specified $g_D=4$, and $m_n\simeq m_p$, $m_D\simeq2m_p$, gives
$$
\boxed{\frac{n_D}{n_nn_p}\simeq\left(\frac{\pi m_pkT}{h^2}\right)^{-3/2}e^{B_D/kT}.}
$$
For $X_a=n_a/n_B$ and $n_B=\eta n_\gamma$, the corresponding <deuterium equilibrium abundance> is
$$
\boxed{\frac{X_D}{X_nX_p}\simeq\frac{16\zeta(3)}{\sqrt\pi}\,\eta\left(\frac{kT}{m_pc^2}\right)^{3/2}e^{B_D/kT}.}
$$
The very small <baryon-to-photon ratio> means that even the high-energy tail contains many photons per baryon capable of destroying deuterium when $kT$ is of order $B_D$. Substantial deuterium survives only after that tail has been exponentially suppressed, so $B_D/kT$ must be large enough to offset the small prefactor. This is the <deuterium bottleneck>, not a threshold set just by equality of thermal <energy> and binding <energy>. The degeneracy $g_D=4$ is used as stipulated for this calculation.
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