Two coherent beams, each of intensity 8.0 W/m² when alone, meet on a screen…

Step 1. At $60^\circ$. $I = 4I_0\cos^2 30^\circ = 4 \times 8.0\ \text{W/m}^2 \times 0.75 = 24.0\ \text{W/m}^2$.

Step 2. At $90^\circ$. $I = 4I_0\cos^2 45^\circ = 4 \times 8.0\ \text{W/m}^2 \times 0.50 = 16.0\ \text{W/m}^2$. This is exactly $2I_0$, the sum of the two beams.

Step 3. At $120^\circ$. $I = 4I_0\cos^2 60^\circ = 4 \times 8.0\ \text{W/m}^2 \times 0.25 = 8.0\ \text{W/m}^2$.

Step 4. The average. Over a whole cycle of the phase, the average of $\cos^2(\varphi/2)$ is $\tfrac12$. So $\bar I = 4I_0 \times \tfrac12 = 2I_0 = 16.0\ \text{W/m}^2$.