Answer
469 V/m.
Work Step by Step
Calculate the intensity of sunlight at Mars. The intensity is the power per unit area. Mars is 1.52 times farther from the sun compared to the Earth. The sun’s power is spread out over a sphere that is $1.52^2$ times larger, so sunlight’s intensity at Mars is reduced by a factor of $1.52^2$ compared to its value at the Earth.
$$\overline{I}_{Mars}=\frac{\overline{I}_{Earth}}{1.52^2}=\frac{1350W/m^2}{1.52^2}\approx584\frac{W}{m^2}$$
The energy per unit area per unit time is the magnitude of the average intensity, equation 22–8. Find the rms value of the electric field.
$$\overline{I}=\frac{1}{2}\epsilon_ocE_o^2=\frac{1}{2}\epsilon_oc(\sqrt{2}E_{rms})^2=\epsilon_ocE_{rms}^2$$
$$E_{rms}=\sqrt{\frac{\overline{I}}{\epsilon_oc }}$$
$$=\sqrt{\frac{584.31W/m^2}{(8.85\times10^{-12} C^2/(N\cdot m^2))(3.00\times10^8m/s)}}$$
$$=469V/m$$
