Answer
$\theta=35.3^{\circ}$
Work Step by Step
Since the light is initially unpolarized, the final intensity will be half of the original intensity. Therefore, $$I_1=\frac{1}{2}I_0$$ Using Malus's law $$I_{out}=I_{in}cos^2\theta$$ Since $I_{out}$ needs to be a third of the original intensity, $I_2=\frac{1}{3}I_o$. Using the value of $I_{in}=\frac{1}{2}I_o$ yields $$\frac{1}{3}I_o=\frac{1}{2}I_ocos^2\theta$$ $$\frac{1}{3}=\frac{1}{2}cos^2\theta$$ Solving for $cos\theta$ yields $$cos\theta=\sqrt{\frac{2}{3}}$$ Using an arccos function on each side yields $$\theta=arccos(\sqrt{\frac{2}{3}})=35.3^{\circ}$$
