= Solution
The conventional electron <plasma frequency> is
$$
\omega_{pe}=\sqrt{\frac{n_e e^2}{\varepsilon_0m_e}}.
$$
Use $n_e=10^{25}\ \mathrm{m^{-3}}$, $e=1.602\times10^{-19}\ \mathrm C$, $m_e=9.109\times10^{-31}\ \mathrm{kg}$ and $\varepsilon_0=8.854\times10^{-12}\ \mathrm{F\,m^{-1}}$. Then
$$
\boxed{\omega_{pe}\simeq1.78\times10^{14}\ \mathrm{rad\,s^{-1}},\qquad
f_{pe}=\frac{\omega_{pe}}{2\pi}\simeq2.84\times10^{13}\ \mathrm{Hz}.}
$$
The equal proton density ensures charge neutrality; it does not double the electron density in this formula. Protons have their own <plasma frequency>,
$$
\omega_{pi}=\sqrt{\frac{n_p e^2}{\varepsilon_0m_p}}
\simeq4.16\times10^{12}\ \mathrm{rad\,s^{-1}}.
$$
If both cold species are allowed to move in the relative charge-separation oscillation, its squared frequency is $\omega_{pe}^2+\omega_{pi}^2$. This gives $1.7843\times10^{14}\ \mathrm{rad\,s^{-1}}$, only about $0.027\%$ above the electron-only answer, because $m_p\gg m_e$.
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