JEE Main 2021 — Electrostatics Question with Solution

The electric field in a region is given by $$\overrightarrow{E}=\left(\frac{3}{5}{E}_0\hat{i}+\frac{4}{5}{E}_0\hat{j}\right)\frac{N}{C}$$. The ratio of flux of reported field through the rectangular surface of area 0.2 m² (parallel to y $$-$$ z plane) to that of the surface of area 0.3 m² (parallel to x $$-$$ z plane) is a : b, where a = __________ [Here $$\hat{i}$$, $$\hat{j}$$ and $$\hat{k}$$ are unit vectors along x, y and z-axes respectively.]