d
\(\overrightarrow{ E }=\left(\frac{3 E _{0}}{5} \hat{ i }+\frac{4 E _{0}}{5} \hat{ j }\right) \frac{ N }{ C }\)

\(A_{1}=0.2 m ^{2}\) [parallel to \(y - z\) plane \(]\)

\(=\overrightarrow{ A }_{1}=0.2 m ^{2} \hat{ i }\)

\(A_{2}=0.3 m ^{2}\) [parallel to \(x - z\) plane \(]\)

\(\overrightarrow{ A }_{2}=0.3 m ^{2} \hat{ j }\)

Now \(\phi_{a}=\left[\frac{3 E_{0}}{5} \hat{i}+\frac{4 E_{0}}{5} \hat{j}\right] \cdot[0.2 \hat{i}]=\frac{3 \times 0.2}{5} E _{0}\)

\( \phi_{b}=\left[\frac{3 E_{0}}{5} \hat{i}+\frac{4 E_{0}}{5} \hat{j}\right] \cdot[0.3 \hat{j}]=\frac{4 \times 0.3}{5} E _{0}\)

\(\operatorname{Now} \frac{\phi_{ a }}{\phi_{ b }}=\frac{0.6}{1.2}=\frac{1}{2}=\frac{ a }{ b }\)

\(\Rightarrow a: b=1: 2\)

\(\Rightarrow a=1\)