= Solution
Because $X_i=n_i/n_B$,
$$
\frac{X_D}{X_pX_n}=n_B\frac{n_D}{n_pn_n}.
$$
Using $n_B=\eta n_\gamma$ and the given photon density in the <deuterium equilibrium abundance> formula gives
$$
\boxed{
\frac{X_D}{X_pX_n}
\simeq\frac{16\zeta(3)}{\sqrt\pi}\eta
\left(\frac{k_BT}{m_pc^2}\right)^{3/2}
e^{B_D/(k_BT)}}.
$$
Although the exponential favors deuterium once $k_BT<B_D$, the very small factor $\eta$ expresses the enormous excess of photons. The high-energy tail therefore photodissociates newly formed deuterium until the <temperature> is well below $B_D/k_B$; this is the <deuterium bottleneck>.
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