= Solution
In <chemical equilibrium>, the reaction $p+e\leftrightarrow H+\gamma$ obeys <chemical-potential balance for ionization>
$$
\mu_p+\mu_e=\mu_H+\mu_\gamma=\mu_H,
$$
because a photon gas in equilibrium has <photon chemical potential> $\mu_\gamma=0$. Applying the given nonrelativistic <Maxwell-Boltzmann distribution> to each massive species gives
$$
\frac{n_en_p}{n_H}
=\frac{g_eg_p}{g_H}
\left(\frac{2\pi k_BT}{h^2}\right)^{3/2}
\left(\frac{m_em_p}{m_H}\right)^{3/2}
\exp\left(-\frac{m_e+m_p-m_H}{k_BT}\right).
$$
The <Hydrogen binding energy> is $I=m_e+m_p-m_H$ in units with $c=1$. Thus <Saha ionization equation>[Saha's equation] is
$$
\boxed{
\frac{n_en_p}{n_H}
=\frac{g_eg_p}{g_H}
\left(\frac{2\pi k_BT}{h^2}\right)^{3/2}
\left(\frac{m_em_p}{m_H}\right)^{3/2}e^{-I/(k_BT)}.}
$$
For ground-state hydrogen, $g_e=g_p=2$ and $g_H=4$, so the degeneracy factor is one. Since $m_p\simeq m_H$, the translational factor is commonly written $(2\pi m_ek_BT/h^2)^{3/2}$.
Charge neutrality gives $n_e=n_p$. Write $n_B=n_e+n_H\simeq\eta n_\gamma$. Then
$$
n_e=X_en_B,
\qquad n_H=(1-X_e)n_B,
$$
and hence
$$
\frac{n_en_p}{n_H}=\frac{X_e^2}{1-X_e}n_B.
$$
Combining this with the stated photon density produces
$$
\boxed{
\frac{1-X_e}{X_e^2}
=\eta\,\frac{g_H}{g_eg_p}
\frac{16\pi\zeta(3)}{(2\pi)^{3/2}}
(k_BT)^{3/2}
\left(\frac{m_H}{m_em_p}\right)^{3/2}
e^{I/(k_BT)}.}
$$
In our universe the <baryon-to-photon ratio> is tiny, so there are roughly $10^9$ photons per baryon. Even when the mean photon energy is far below $I$, the high-energy <blackbody radiation> tail contains enough ionizing photons to suppress neutral hydrogen. The <recombination temperature> is reached only after the <Boltzmann factor> overwhelms this large photon-to-baryon ratio, at $k_BT\simeq0.3\,\mathrm{eV}$.
For a universe with $\eta=1$, take the degeneracy factor as one, $m_H\simeq m_p$, and define $x=I/(k_BT)$. Neutral formation occurs when the boxed ratio is of order one, giving approximately
$$
1\simeq\frac{4\sqrt2\,\zeta(3)}{\sqrt\pi}
\left(\frac{I}{m_ex}\right)^{3/2}e^x.
$$
With $I=13.6\,\mathrm{eV}$ and $m_e=5.11\times10^5\,\mathrm{eV}$, this gives $x\simeq19$ and therefore
$$
\boxed{k_BT\simeq0.7\,\mathrm{eV},}
$$
to order unity. This is much hotter than recombination in our universe because there is no enormous excess of photons, but it remains below $I$ because the electron's large translational phase space still favours the ionized state.
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