Given field

The electric field is uniform and along the $x$-direction: $\vec E = E_0\hat x$

Flux through a surface

Electric flux through a plane surface is $\Phi = \vec E \cdot \vec A = EA\cos\theta$ where $\theta$ is the angle between the electric field and the area vector (normal to the surface).

Using projection idea

Since the field is along $x$-direction, the flux through any surface equals $\Phi = E_0 \times (\text{projection of the area on the plane perpendicular to } x)$

The plane perpendicular to $x$ is the $yz$-plane.

From the figure

The shaded area is an inclined square/plane of side $a$, so its actual area is not what directly matters; what matters is its projection on the $yz$-plane.

From the geometry of the figure, the projected area on the plane normal to $\hat x$ is $a^2$

Compute the flux

Therefore, $\Phi = E_0 \cdot a^2$

Match with options

This corresponds to: $\boxed{\text{C: } E_0 a^2}$

Verification with stored answer

Stored correct answer = C, which matches our derived result.