= Solution
Put $x=E_I/(k_BT)$. With $\zeta=1/2$, the <heat capacity> formula becomes
$$
50=\frac94+\frac16\left(\frac32+x\right)^2,
\quad x=\sqrt{286.5}-\frac32=15.4263.
$$
Taking $E_I=13.6\,\mathrm{eV}$ gives $\boxed{T=1.02\times10^4\,\mathrm K}$. At half <ionization>, $n_en_p/n_H=n_e$, so the <Saha ionization equation> directly gives
$$
\boxed{n_e=\left(\frac{2\pi m_ek_BT}{h^2}\right)^{3/2}e^{-x}\simeq5.0\times10^{20}\,\mathrm{m^{-3}}.}
$$
This uses $2g_p/g_H=1$ as specified in part (a); a different internal-state convention must be carried consistently into the numerical <electron density>. The original nuclei <density> is $n_0=2n_e\simeq10^{21}\,\mathrm{m^{-3}}$.
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