= Solution
For cold, collisionless, unmagnetized electrons with stationary heavy ions, $-m_e\omega^2\mathbf x=-e\mathbf E$. Therefore $\mathbf P=-ne\mathbf x=-ne^2\mathbf E/(m_e\omega^2)$ and
$$
\boxed{\epsilon_r(\omega)=1-\frac{\omega_p^2}{\omega^2},\qquad
\omega_p=\sqrt{\frac{ne^2}{m_e\epsilon_0}}.}
$$
This is the <cold collisionless plasma dielectric response>. For a propagating transverse wave, $\mathbf B=\mathbf k\times\mathbf E/\omega$, so $E/B=c$ in vacuum and $E/B=c/n_p$ in <plasma>, where $n_p=\sqrt{\epsilon_r}$. Equivalently $E/H=Z_0$ and $Z_0/n_p$ for $\mu=\mu_0$.
Converting the <density> gives $n=10^{18}\,\mathrm{m^{-3}}$, so $\omega_p=5.64\times10^{10}\,\mathrm{s^{-1}}$ and $f_p=8.98\,\mathrm{GHz}$. At $10\,\mathrm{GHz}$,
$$
\boxed{\epsilon_r\simeq0.19384,\qquad n_p\simeq0.4403.}
$$
At an interface without independently imposed surface charge or sheet current, tangential $\mathbf E,\mathbf H$ and normal $\mathbf B,\mathbf D$ are continuous, with the electron response included in the effective <permittivity>. Normal $E$ need not be continuous; the induced electron charge is represented by that response. In a convention treating it explicitly as a surface charge, the normal vacuum-displacement jump equals that charge instead.
For the downward transmitted wave, transversality gives $k_xE_x+k_zE_z=0$. The measured field therefore has $k_x/(-k_z)=3$, so $\tan\theta_t=3$ and $\sin\theta_t=3/\sqrt{10}$. Tangential <phase matching> gives
$$
\boxed{\theta_i=\sin^{-1}\left(n_p\frac3{\sqrt{10}}\right)\simeq24.69^\circ.}
$$
For <inferring incident power from a transmitted plasma field>, choose $\mathbf E_i=E_i(\cos\theta_i,0,\sin\theta_i)$ and $\mathbf E_r=E_r(\cos\theta_i,0,-\sin\theta_i)$. Their <magnetic fields> have opposite $y$ directions. The two tangential conditions are
$$
(E_i+E_r)\cos\theta_i=1\,\mathrm{V\,m^{-1}},\qquad
E_i-E_r=n_p\sqrt{10}\,\mathrm{V\,m^{-1}}.
$$
Hence $E_i\simeq1.2464\,\mathrm{V\,m^{-1}}$ and $E_r\simeq-0.1458\,\mathrm{V\,m^{-1}}$. Since these are peak phasor amplitudes,
$$
\boxed{\langle S_i\rangle=\frac{E_i^2}{2Z_0}
\simeq2.06\times10^{-3}\,\mathrm{W\,m^{-2}}.}
$$
This is incident <power> per area perpendicular to its propagation. Per area of the interface the incoming normal flux is $\langle S_i\rangle\cos\theta_i\simeq1.87\times10^{-3}\,\mathrm{W\,m^{-2}}$. The reflected sign is a phase reversal, not negative reflected <energy>.
At $1\,\mathrm{GHz}<f_p$, the <relative permittivity> is negative and there is no propagating transverse bulk wave. The field decays evanescently into the <plasma> and the ideal collisionless half-space reflects all incoming normal <power>. Collisions would allow absorption; a <complex refractive index> here does not by itself imply dissipative heating.
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