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(D) The electric field is $\overrightarrow{E} = \frac{3 E_{0}}{5} \hat{i} + \frac{4 E_{0}}{5} \hat{j} \, N/C$.For the first surface,the area vector is

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(D) The electric field is $\overrightarrow{E} = \frac{3 E_{0}}{5} \hat{i} + \frac{4 E_{0}}{5} \hat{j} \, N/C$.For the first surface,the area vector is $\overrightarrow{A}_{1} = 0.2 \hat{i} \, m^{2}$ (since it is parallel to the $y-z$ plane).The flux $\phi_{1} = \overrightarrow{E} \cdot \overrightarrow{A}_{1} = (\frac{3 E_{0}}{5} \hat{i} + \frac{4 E_{0}}{5} \hat{j}) \cdot (0.2 \hat{i}) = \frac{3 	imes 0.2}{5} E_{0} = 0.12 E_{0} \, V \cdot m$.For the second surface,the area vector is $\overrightarrow{A}_{2} = 0.3 \hat{j} \, m^{2}$ (since it is parallel to the $x-z$ plane).The flux $\phi_{2} = \overrightarrow{E} \cdot \overrightarrow{A}_{2} = (\frac{3 E_{0}}{5} \hat{i} + \frac{4 E_{0}}{5} \hat{j}) \cdot (0.3 \hat{j}) = \frac{4 	imes 0.3}{5} E_{0} = 0.24 E_{0} \, V \cdot m$.The ratio of the fluxes is $\frac{\phi_{1}}{\phi_{2}} = \frac{0.12 E_{0}}{0.24 E_{0}} = \frac{1}{2}$.Given the ratio is $a : b = 1 : 2$,therefore $a = 1$.

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