= Solution
Assume a dilute nondegenerate <hydrogen> <gas> in thermal and <chemical equilibrium>, with Maxwell-Boltzmann <translational partition functions>, ground-state neutral <atoms>, single <ionization> and <charge neutrality> $n_p=n_e$. For $\mathrm H\rightleftharpoons p+e$, <chemical equilibrium> gives $\mu_H=\mu_p+\mu_e$. The ideal-gas relation $\mu_i=E_{i0}+k_BT\log(n_i\lambda_i^3/g_i)$, with $E_{H0}=-R$, gives
$$
\frac{n_pn_e}{n_H}=\frac{g_pg_e}{g_H}\frac{\lambda_H^3}{\lambda_p^3\lambda_e^3}e^{-R/k_BT}.
$$
Neglecting the <electron> <mass> in $m_H/m_p$ and using the ground-state electronic statistical weights, $g_e=2$ and $g_H=2$ with cancelling nuclear-spin factors, this becomes
$$
\boxed{\frac{n_e^2}{n_H}=S(T)=\frac{(2\pi m_ek_BT)^{3/2}}{h^3}e^{-R/k_BT}.}
$$
Excited neutral states have an <energy> gap of at least $3R/4$ above the ground state. At the <temperatures> relevant to the plotted maximum their <Boltzmann factors> are very small despite the finite degeneracy increase. Highly excited levels also require a physical <plasma> cutoff; the untruncated infinite isolated-atom partition sum is not a valid finite-density model. Ground-state dominance is an approximation in this regime, not a theorem at all <temperatures>.
Take the heating process to keep the total hydrogen-nucleus <density> fixed, so $n_H+n_e=n_0$. This fixed-density assumption is needed to determine $n_0$ from the curve; freely expanding <gas> would require its <density> history. The excited-to-ground ratio is $n_2/n_H\simeq4e^{-3R/(4k_BT)}$. Therefore
$$
\boxed{f=\frac{n_2}{n_0}\simeq4(1-x)e^{-3R/(4k_BT)}
=\frac4{1+n_e/n_H}e^{-3R/(4k_BT)},\qquad x=n_e/n_0.}
$$
At low <temperature> excitation is suppressed. At higher <temperature> excitation grows, but eventually <ionization> removes neutral <atoms> and the excited-neutral abundance decreases. To locate the maximum quantitatively, write $u=R/(k_BT)$. From $S/n_0=x^2/(1-x)$,
$$
\frac{dx}{d\log T}=\frac{x(1-x)}{2-x}\left(\frac32+u\right).
$$
Setting $d\log f/d\log T=0$ gives
$$
\boxed{x_{\max}=\frac{6u}{6+7u}.}
$$
Reading the original graph gives $T_{\max}\simeq(7.9\text{–}8.0)\times10^3$ K and $f_{\max}\simeq2.3\times10^{-7}$. With $R=13.6$ eV, this implies $u\simeq20$ and $\boxed{x_{\max}\simeq0.82}$. Substitution into
$$
\boxed{n_0=S(T_{\max})\frac{1-x_{\max}}{x_{\max}^2}}
$$
gives approximately $10^{18}\,\mathrm{m^{-3}}$. The result is particularly temperature-sensitive because of the Saha exponential; using $7900$ K gives about $9\times10^{17}\,\mathrm{m^{-3}}$, while the coarse $8000$ K reading gives about $1.2\times10^{18}\,\mathrm{m^{-3}}$. The plotted height is consistent with this range and should not be used to claim more precision than the source graph provides.
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