The electric field of a plane electromagnetic wave propagating in a...

The amplitude of the electric field is $E_0 = 60\pi$ V/m.

The average intensity of an electromagnetic wave in a medium is given by $I = \dfrac{1}{2} \epsilon v E_0^2$, where $\epsilon$ is the permittivity of the medium and $v$ is the speed of the wave.

For a non-magnetic medium, the refractive index is $n = \sqrt{\epsilon_r}$, so $\epsilon = n^2 \epsilon_0$. The speed of the wave is $v = \dfrac{c}{n}$.

Substituting these into the intensity formula gives:
$I = \dfrac{1}{2} (n^2 \epsilon_0) \left(\dfrac{c}{n}\right) E_0^2 = \dfrac{1}{2} n \epsilon_0 c E_0^2$

We know that $\epsilon_0 c = \dfrac{1}{\mu_0 c} = \dfrac{1}{120\pi} \Omega^{-1}$.

Substituting the values $n = 3$ and $E_0 = 60\pi$ V/m:
$I = \dfrac{1}{2} \times 3 \times \dfrac{1}{120\pi} \times (60\pi)^2$

$I = \dfrac{3}{240\pi} \times 3600\pi^2 = 45\pi$ W/m$^2$.

Answer: $45\pi$ W/m$^2$