= Solution
The <ion> <density> is $\rho/(9m_u)$ and double ionization gives $n_e\simeq2.48\times10^{29}$ m$^{-3}$. Therefore $\hbar\omega_p\simeq18.5$ eV. The <photon energy> loss gives $\hbar\omega=30$ eV, while <momentum conservation> gives
$$
k=\frac{\sqrt{E_i^2+E_f^2-2E_iE_f\cos(\pi/4)}}{\hbar c}\simeq1.14\times10^{10}\ \mathrm{m^{-1}}.
$$
Inverting the stipulated dispersion law gives
$$
\boxed{k_BT=\frac{m_e(\omega^2-\omega_p^2)}{3k^2}\simeq18.7\ \mathrm{eV}.}
$$
An <angular frequency> expressed “in eV” means $\hbar\omega$, not the <frequency> itself. The <temperature> is a formal inference conditional on the dispersion model, whose validity is tested next.
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