= Solution
Define fractional ionization as $x=n_p/(n_H+n_p)$ and use electrical neutrality. For $x=0.01$ and $n_e=10^9\,\mathrm{m^{-3}}$, $n_H=(1-x)n_e/x=99n_e$. Thus the density form of the <Saha equation> requires
$$
\left(\frac{2\pi m_ek_{\rm B}T}{h^2}\right)^{3/2}e^{-13.6\,\mathrm{eV}/(k_{\rm B}T)}=\frac{n_e}{99}\simeq1.01\times10^7\,\mathrm{m^{-3}}.
$$
Solving this monotone equation, conveniently after taking its logarithm, gives
$$
\boxed{T\simeq3.48\times10^3\,\mathrm K,\qquad k_{\rm B}T\simeq0.300\,\mathrm{eV}.}
$$
The thermal energy is only about $2.2\%$ of the $13.6$ eV ionization energy. Ionization is not decided by setting $k_{\rm B}T=R$: the high-energy tail and the large translational state-count factor in chemical equilibrium allow a residual ionized fraction at a much lower temperature. This is the equilibrium estimate for the stipulated density and fraction, without inferring an expansion age from it.
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