= Solution
For a nonrelativistic, three-dimensional ideal electron gas with spin degeneracy two,
$$
E_F=\frac{\hbar^2}{2m_e}(3\pi^2n_e)^{2/3},\qquad \omega_{\rm pe}=\sqrt{\frac{n_ee^2}{m_e\epsilon_0}}.
$$
Dividing gives
$$
\boxed{\frac{\hbar\omega_{\rm pe}}{E_F}=\sqrt{\frac{4e^2m_e}{\hbar^2\epsilon_0}}(3\pi^2)^{-2/3}n_e^{-1/6}.}
$$
Write the density-independent coefficient as $K$. From $\hbar\omega_{\rm pe}/E_F=1.384$, $n_e=(K/1.384)^6$. In this free-electron model,
$$
\boxed{n_e\simeq1.59\times10^{29}\,\mathrm{m^{-3}},\qquad E_F\simeq10.69\,\mathrm{eV},\qquad \hbar\omega_{\rm pe}\simeq14.79\,\mathrm{eV}.}
$$
The result uses the stipulated energy ratio, rather than substituting a separately tabulated aluminium density or plasmon energy.
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