The electric field in a region is given by E → = 3 5 E 0 i ^ + 4 5 E 0 j ^   N   C - 1 . The…

$\overrightarrow{E}=\left(\frac{3E_0}{5}\hat{\mathrm{i}}+\frac{4E_0}{5}\hat{\mathrm{j}}\right) \mathrm{N} {\mathrm{C}}^{-1}$

$A_1=0.2 {\mathrm{m}}^2$ [parallel to $y-z$ plane]

$={\overrightarrow{A}}_1=0.2 {\mathrm{m}}^2\hat{\mathrm{i}}$

$A_2=0.3 {\mathrm{m}}^2$ [parallel to $x-z$ plane]

${\overrightarrow{\mathrm{A}}}_2=0.3 {\mathrm{m}}^2\hat{\mathrm{j}}$

Now, ${\varphi }_a=\left[\frac{3E_0}{5}\hat{\mathrm{i}}+\frac{4E_0}{5}\hat{\mathrm{j}}\right]·\left[0.2\hat{\mathrm{i}}\right]=\frac{3\times 0.2}{5}{E}_0$

$\& {\varphi }_{\mathrm{b}}=\left[\frac{3E_0}{5}\hat{\mathrm{i}}+\frac{4E_0}{5}\hat{\mathrm{j}}\right]·\left[0.3\hat{\mathrm{j}}\right]=\frac{4\times 0.3}{5}{E}_0$

Now, $\frac{{\varphi }_a}{{\varphi }_b}=\frac{0.6}{1.2}=\frac{1}{2}=\frac{a}{b}$

$\Rightarrow a:b=1:2$

$\Rightarrow a=1$