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+### Statement
+
+$12.1.18.$ [Insert the problem statement]
+
+### Solution
+
+To estimate the penetration depth of an electromagnetic wave, the free electron plasma model is used.
+
+At optical frequencies, the permittivity of the conductor is
+
+$\varepsilon(\omega) = 1 - \frac{\omega_p^2}{\omega^2},
+\qquad \omega = 2\pi\nu$,
+
+where the plasma frequency in SI is
+
+$\omega_p = \sqrt{\frac{n_e e^2}{\varepsilon_0 m_e}}$.
+
+With e = $1.6\times10^{-19}\,\text{C}, \varepsilon_0 = 8.85\times10^{-12}\,\text{F/m}, m_e = 9.1\times10^{-31}\,\text{kg}$:
+
+$\omega_p^2 = \frac{10^{28}(1.6\times10^{-19})^2}{8.85\times10^{-12}\times9.1\times10^{-31}} \approx 3.2\times10^{31}\,\text{rad}^2/\text{s}^2,
+\quad \omega_p \approx 5.6\times10^{15}\,\text{rad/s}$
+
+$Since \omega = 2\pi\times10^{15} \approx 6.3\times10^{15}\,\text{rad/s} > \omega_p$
+the wave propagates with attenuation. The penetration depth \delta in a conductor is given by the damping length of the field intensity:
+
+$\delta = \frac{c}{\sqrt{\omega_p^2 - \omega^2}} \approx \frac{c}{\omega_p}
+\quad (\text{since } \omega \sim \omega_p)$
+
+Numerically,
+
+$\delta \approx \frac{3\times10^{10}\,\text{cm/s}}{5.6\times10^{15}\,\text{s}^{-1}} \approx 5.4\times10^{-6}\,\text{cm}$
+
+$\boxed{\delta \approx 5\times10^{-6}\,\text{cm} \; (= 50\,\text{nm})}$.
+
+#### Answer
+
+[Insert a concise answer or boxed result]