Answer
$\theta = 20^{\circ}, 70^{\circ}$
Work Step by Step
We can write an expression for $I_1$:
$I_1 = I_0~cos^2~\theta$
We can write an expression for $I_2$:
$I_2 = I_1~cos^2~(90^{\circ}-\theta)$
$I_2 = ( I_0~cos^2~\theta)~cos^2~(90^{\circ}-\theta)$
$I_2 = I_0~cos^2~\theta~sin^2~\theta$
We can find $\theta$:
$I_2 = I_0~cos^2~\theta~sin^2~\theta$
$I_2 = I_0~(\frac{1}{4}~sin^2~2\theta) = 0.10~I_0$
$sin^2~2\theta = 0.40$
$sin~2\theta = 0.63246$
$2\theta = sin^{-1}~(0.63246)~~$ or $~~2\theta = 180^{\circ}-sin^{-1}~(0.63246)$
$2\theta = 39.2^{\circ}~~$ or $~~2\theta = 140.8^{\circ}$
$\theta = 20^{\circ}, 70^{\circ}$
