In a certain region, the electric field is expressed as E = (3)/(5)...

The surface is parallel to the $y$-$z$ plane, so its area vector points along the $x$-axis.

$\vec{A} = 0.2 \hat{i} \text{ m}^2$

The electric field is given as:

$\vec{E} = \dfrac{3}{5} E_0 \hat{i} + \dfrac{4}{5} E_0 \hat{j}$

The electric flux $\Phi$ through the surface is the dot product of the electric field and the area vector:

$\Phi = \vec{E} \cdot \vec{A}$

$\Phi = \left( \dfrac{3}{5} E_0 \hat{i} + \dfrac{4}{5} E_0 \hat{j} \right) \cdot (0.2 \hat{i})$

$\Phi = \dfrac{3}{5} E_0 \times 0.2$

Substituting $E_0 = 2.0 \times 10^3 \text{ N C}^{-1}$:

$\Phi = \dfrac{3}{5} \times 2000 \times 0.2$

$\Phi = 1200 \times 0.2 = 240 \text{ N m}^2 \text{ C}^{-1}$