Answer
$\theta = cos^{-1}~\sqrt{\frac{p}{50}}$
Work Step by Step
Let $I_0$ be the original intensity of the light.
Since the light is initially unpolarized, half the intensity will be transmitted through the first polarizing sheet.
$I_1 = \frac{1}{2}I_0$
We can find an expression for $I_2$:
$I_2 = I_1~cos^2~\theta^{\circ}$
$I_2 = \frac{1}{2}I_0~cos^2~\theta$
We can find $\theta$:
$I_2 = \frac{1}{2}I_0~cos^2~\theta = \frac{p}{100}~I_0$
$cos^2~\theta = \frac{p}{50}$
$cos~\theta = \sqrt{\frac{p}{50}}$
$\theta = cos^{-1}~\sqrt{\frac{p}{50}}$
