As electric field is in $y$-direction so electric flux is only due to top and bottom surface Bottom surface $y=0$ $$ \Rightarrow E=0 \Rightarrow \varphi=0 $$ Top surface $\mathrm{y}=0.5 \mathrm{~m}$ $$ \Rightarrow E=150(.5)^2=\frac{150}{4} $$ Now flux $\phi=\mathrm{EA}=\frac{150}{4}(0.5)^2=\frac{150}{16}$ By Gauss's law $$ \phi=\frac{Q_{i n}}{\epsilon_0} $$ $$ \begin{aligned} & \frac{150}{16}=\frac{Q_n}{\epsilon_0} \\ & Q_{\text {in }}=\frac{150}{16} \times 8.85 \times 10^{-12}=8.3 \times 10^{-11} \mathrm{C} \end{aligned} $$ Option (2)