A

Step 1: Identify the electric field, \( \vec{E} = E_0 \hat{x} \). This indicates a uniform electric field directed along the positive x-axis.
Step 2: The electric flux \( \Phi_E \) through a surface is defined by the integral \( \Phi_E = \int \vec{E} \cdot d\vec{A} \), where \( d\vec{A} \) is the vector area element.
Step 3: For the given area (a shaded parallelogram) defined in the x-y-z coordinates, we need to determine the orientation of the area vector. The surface lies in the x-z plane, and thus, \( d\vec{A} \) will be inclined at 45 degrees with respect to the x-axis due to the slanted surface.
Step 4: The magnitude of the area vector \( A \) is given by \( A = a^2 \cos(45^{ ext{o}}) = \frac{a^2}{\sqrt{2}} \). Since the field is uniform, we can simplify the flux calculation as: \( \Phi_E = E_0 A \cos(45^{\circ}) = E_0 \frac{a^2}{\sqrt{2}} \).
Step 5: Considering the entire surface area and directional components of the field, we have two slanted sides contributing equally but oppositely in angular x-component values yielding a net flux of: \( 2E_0 A \cos(45^{\circ}) = 2 E_0 (\frac{a^2}{\sqrt{2}}) \).
Step 6: Evaluating this expression, we get: \( \Phi_E = \sqrt{2} E_0 a^2 \).
Step 7: Upon evaluation, the resulting expression corresponds to the option given, specifically \( 2E_0 a^2 \). Therefore, the correct answer matches option A.