A hollow hemispherical bowl of radius 0.16 m has its rim in a horizontal plane and its curved part…

Step 1. The opening. Its area is $A = \pi r^2 = \pi \times (0.16\ \text{m})^2 = 0.0804\ \text{m}^2$. The outward normal points up, along the field, so $\theta = 0^\circ$. $\Phi = EA = 350\ \text{N/C} \times 0.0804\ \text{m}^2 = 28.1\ \text{N m}^2/\text{C}$.

Step 2. The net flux. The field is uniform and the surface is closed, so $\Phi_{\text{net}} = 0$.

Step 3. The curved surface. $\Phi_{\text{curved}} = 0 - 28.1 = -28.1\ \text{N m}^2/\text{C}$. It is negative, because the lines enter the bowl through its curved wall and leave through the opening.