= Solution
With $X_e=n_e/n_b$, charge neutrality and $n_b=n_p+n_H$ give
$$
n_e=X_en_b,
\qquad n_H=(1-X_e)n_b.
$$
Multiplying part a by $n_b=\eta n_\gamma$ and using $n_\gamma=2\zeta(3)T^3/\pi^2$ yields the <Saha ionization equation>
$$
\boxed{\left(\frac{1-X_e}{X_e^2}\right)_{\rm eq}
=\frac{2\zeta(3)}{\pi^2}\eta
\left(\frac{2\pi T}{m_e}\right)^{3/2}e^{B_H/T}}.
$$
Although $T\ll B_H$ suppresses ionization of an individual atom, the <baryon-to-photon ratio> is only about $10^{-9}$. The enormous number of photons per baryon leaves enough photons in the high-energy thermal tail to ionize hydrogen until $T\simeq0.3\,\mathrm{eV}$, far below $B_H$.
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