Answer
$\theta_2 = 70.4^{\circ}$
Work Step by Step
Since the light is initially unpolarized, half the intensity will be transmitted through the first polarizing sheet.
We can write an expression for $I_1$:
$I_1 = \frac{1}{2}I_0$
We can write an expression for $I_2$:
$I_2 = I_1~cos^2~\theta_2$
$I_2 = \frac{1}{2}I_0~cos^2~\theta_2$
We can write an expression for $I_3$:
$I_3 = I_2~cos^2~(90^{\circ}-\theta_2)$
$I_3 = \frac{1}{2}I_0~cos^2~\theta_2~cos^2~(90^{\circ}-\theta_2)$
$I_3 = \frac{1}{2}I_0~cos^2~\theta_2~sin^2~\theta_2$
We can find $\theta_2$:
$I_3 = \frac{1}{2}I_0~cos^2~\theta_2~sin^2~\theta_2$
$I_3 = I_0~(\frac{1}{8}~sin^2~2\theta_2) = 0.0500~I_0$
$sin^2~2\theta_2 = 0.400$
$sin~2\theta_2 = 0.63246$
$2\theta_2 = sin^{-1}~(0.63246)~~$ or $~~2\theta_2 = 180^{\circ}-sin^{-1}~(0.63246)$
$2\theta_2 = 39.2^{\circ}~~$ or $~~2\theta_2 = 140.8^{\circ}$
$\theta_2 = 19.6^{\circ}, 70.4^{\circ}$
