Unpolarised light of intensity 150 W/m² meets a Polaroid…

Step 1. The intensity. $I = \dfrac{I_0}{2}\cos^2\theta = \dfrac{150\ \text{W/m}^2}{2} \times \cos^2 60^\circ = 75\ \text{W/m}^2 \times 0.25 = 18.75\ \text{W/m}^2$.

Step 2. The fraction. $\dfrac{I}{I_0} = \dfrac{18.75\ \text{W/m}^2}{150\ \text{W/m}^2} = 0.125$, which is $\tfrac12\cos^2 60^\circ$.

Step 3. The angle. We need $\tfrac12\cos^2\theta = 0.10$, so $\cos^2\theta = 0.20$ and $\cos\theta = 0.447$. Then $\theta = 63.4^\circ$.