= Solution
Write $n_b=n_e+n_H\simeq\eta n_\gamma$. Charge neutrality gives $n_p=n_e$, and therefore
$$
n_e=X_en_b,
\qquad
n_H=(1-X_e)n_b.
$$
Substitution into the <Saha ionization equation> yields
$$
\frac{1-X_e}{X_e^2}
=n_b\left(\frac{2\pi\hbar^2}{m_ek_BT}\right)^{3/2}
e^{I/(k_BT)}.
$$
Using the given photon density,
$$
\boxed{
\frac{1-X_e}{X_e^2}
=\eta\frac{2\zeta(3)(2\pi)^{3/2}}{\pi^2}
\left(\frac{k_BT}{m_ec^2}\right)^{3/2}
e^{I/(k_BT)}}.
$$
The <baryon-to-photon ratio> is extremely small. Even when $k_BT\ll I$, the high-energy tail of the enormous photon population contains enough ionizing photons per baryon to delay <cosmological recombination>. The exponential factor must become large enough to overcome both $\eta\sim10^{-9}$ and the phase-space factor, which occurs near $k_BT\simeq0.3\,\mathrm{eV}$ rather than near $I$.
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