Explanation

(a) Using Malus law, $I = I_{0}\cos^{2}\theta$
As here polarizer is rotating, $i.e.,$ all the values of $\theta$ are possible
$I_{av} = \frac{1}{2\pi}\int_{0}^{2\pi}{I\ d\theta} = \frac{1}{2\pi}\int_{0}^{2\pi}I_{0}\cos^{2}\theta\ d\theta$
On integration we get $I_{av} = \frac{I_{0}}{2}$
Where $I_{0} = \frac{Energy}{Area \times Time} = \frac{p}{A} = \frac{10^{- 3}}{3 \times 10^{- 4}} = \frac{10}{3}\frac{Watt}{m^{2}}$
$\therefore I_{av} = \frac{1}{2} \times \frac{10}{3} = \frac{5}{3}Watt$
and Time period $T = \frac{2\pi}{\omega} = \frac{2 \times 3.14}{31.4} = \frac{1}{5}s$
$\therefore$ Energy of light passing through the polarizer per revolution $= I_{av} \times Area \times T = \frac{5}{3} \times 3 \times 10^{- 4} \times \frac{1}{5} = 10^{- 4}J$