= Solution
Assume an ideal nondegenerate hydrogen gas in <thermal equilibrium> and <chemical equilibrium>, with ground-state atoms and charge neutrality. Let $n_p=n_e=Xn_B$ and $n_H=(1-X)n_B$. The ground-state <Saha equation>, with the usual electron/proton spin factors cancelling the atomic ground-state degeneracy, is
$$
\frac{n_en_p}{n_H}=\left(\frac{m_ek_BT}{2\pi\hbar^2}\right)^{3/2}e^{-B/(k_BT)}.
$$
Invert it and use $n_B=n_\gamma/s=(0.24/s)(k_BT/(\hbar c))^3$. This gives
$$
\frac{1-X}{X^2}=\frac{0.24(2\pi)^{3/2}}s\left(\frac{k_BT}{m_ec^2}\right)^{3/2}e^{B/(k_BT)},
$$
so
$$
\boxed{\frac{1-X}{X^2}\simeq\frac{3.8}s\left(\frac{k_BT}{m_ec^2}\right)^{3/2}\exp\frac{B}{k_BT}.}
$$
Here $m_e$ is the electron mass denoted $M_e$ in the source, and $0.24(2\pi)^{3/2}=3.78$. At $X=1/2$, the left side is two. At $k_BT$ of order $B$, the right side is extremely small because $1/s\sim10^{-9}$ and $B/(m_ec^2)\ll1$. A large exponential is needed to compensate, requiring $B/(k_BT)\gg1$ rather than a temperature of order the binding energy.
Physically, there are enormously many photons per baryon, so even a small high-energy tail of the thermal photon distribution can maintain ionization. Recombination is delayed until that tail is sufficiently depleted. This explains the low recombination temperature without solving the equilibrium equation; actual cosmological recombination also eventually requires rate equations once equilibrium fails.
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