b
As electric field is in \(y-\)direction so electric flux is only due to top and bottom surface

Bottom surface \({y}=0\)

\(\Rightarrow {E}=0 \Rightarrow \phi=0\)

Top surface \({y}=0.5\, {m}\)

\(\Rightarrow {E}=150(0.5)^{2}=\frac{150}{4}\)

Now flux \(\phi={EA}=\frac{150}{4}(.5)^{2}=\frac{150}{16}\)

By Gauss's law \(\phi=\frac{Q_{\text {in }}}{\epsilon_{0}}\)

\(\frac{150}{16}=\frac{{Q}_{\text {in }}}{\epsilon_{0}}\)

\({Q}_{\text {in }}=\frac{150}{16} \times 8.85 \times 10^{-12}=8.3 \times 10^{-11} \,{C}\)