The parallel equipotential lines represent presence of a uniform Electric field in the region.

Electric field will be E perpendicular to the equipotential surfaces as shown in figure.

Slope of equipotential surfaces,

$\tan \theta=\frac{1-0}{4-2}=\frac{1}{2}$

$\Rightarrow \sin \theta=\frac{1}{\sqrt{5}} \text { and } \cos \theta=\frac{2}{\sqrt{5}}$

Here, potential difference $\Delta V=2$ volt

Along the direction of Electric field.

Distance between two equipotential lines, $d=0.02 \sin \theta=\frac{0.2}{\sqrt{5}}$

Using, $\Delta V = Ed$ we have,

$2=E \times \frac{0.02}{\sqrt{5}}$

$\therefore E =100 \sqrt{5}\,V / m$

From the geometry shown in figure,

$E_x=-E \sin \theta=-100 \sqrt{5} \times \frac{1}{\sqrt{5}}$

$\therefore E_x=-100\,V / m$

similary,

$E_y=E \cos \theta=100 \sqrt{5} \times \frac{2}{\sqrt{5}}$

$\therefore E_y=200\,V / m$

Hence, option $(c)$ is the correct answer.

Why this question?

In such questions the parallel equipotential surface gives hint for uniform $\vec{E}$.

for which we can apply $\Delta V = Ed$ and distance between equipotential surface

$(d)$ in direction of $\vec{E}$ can be calculated by observing geometry.