A closed box measures 0.30 m along x, 0.20 m along y and 0.10 m along z…

Step 1. The area of those two faces. $A = 0.20\ \text{m} \times 0.10\ \text{m} = 0.020\ \text{m}^2$.

Step 2. The face that the field enters. The outward normal points along $-x$, so $\theta = 180^\circ$. $\Phi = -EA = -500\ \text{N/C} \times 0.020\ \text{m}^2 = -10\ \text{N m}^2/\text{C}$.

Step 3. The face that the field leaves. The outward normal points along $+x$, so $\theta = 0^\circ$. $\Phi = +EA = +10\ \text{N m}^2/\text{C}$.

Step 4. The other four faces. The field runs along them. Each has $\theta = 90^\circ$ and zero flux.

Step 5. The net flux. $-10 + 10 + 0 = 0\ \text{N m}^2/\text{C}$.