Solution

To estimate the penetration depth of a high-frequency electromagnetic wave, the free electron plasma model is used.
At optical frequencies, the penetration depth of the field (skin depth) is characterized by the plasma frequency of the conductor $\omega_p$, which in SI is:
$$\omega_p = \sqrt{\frac{n_e e^2}{\varepsilon_0 m_e}}$$

Substitute the known values ($e \approx 1.6 \cdot 10^{-19}\text{ C}$, $\varepsilon_0 \approx 8.85 \cdot 10^{-12}\text{ F/m}$, $m_e \approx 9.1 \cdot 10^{-31}\text{ kg}$) and the electron concentration $n_e = 10^{22}\text{ cm}^{-3} = 10^{28}\text{ m}^{-3}$:
$$\omega_p^2 = \frac{10^{28} \cdot (1.6 \cdot 10^{-19})^2}{8.85 \cdot 10^{-12} \cdot 9.1 \cdot 10^{-31}} \approx 3.18 \cdot 10^{31}\text{ rad}^2/\text{s}^2$$
$$\omega_p \approx 5.6 \cdot 10^{15}\text{ rad/s}$$

The angular frequency of the incident wave is $\omega = 2\pi\nu = 2\pi \cdot 10^{15} \approx 6.3 \cdot 10^{15}\text{ rad/s}$. Since the wave frequency is of the same order of magnitude as the plasma frequency, a rough estimate for the characteristic field attenuation depth $\delta$ in the metal is given by the plasma length:
$$\delta \approx \frac{c}{\omega_p}$$

Numerically:
$$\delta \approx \frac{3 \cdot 10^8\text{ m/s}}{5.6 \cdot 10^{15}\text{ s}^{-1}} \approx 5.4 \cdot 10^{-8}\text{ m}$$
Converting to convenient units, we get $\delta \approx 5 \cdot 10^{-6}\text{ cm}$, which equals $50\text{ nm}$.