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Since the electric field has only an x component, for faces perpendicular to x direction, the angle between E and $\Delta S$ is $\pm \large\frac{\pi}{2}$

Therefore, the flux $\phi = E.\Delta S$ is separately zero for each face of the cube except the two shaded ones.

Now the magnitude of the electric field at the left face is

$E_L = \alpha x^{1/2} = \alpha a^{1/2}$ (x = a at the left face).

The magnitude of electric field at the right face is

$E_R = \alpha x^{1/2} = \alpha (2a)^{1/2} $ (x = 2a at the right face).

The corresponding fluxes are

$\qquad= \alpha a ^{5/2}( \sqrt 2-1)$

$\qquad= 800 (0.1)^{5/2}. (\sqrt 2 -1)$

$\qquad= 1.05 \;Nm^2 C^{-1}$

Hence A is the correct answer.

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