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We can see from the figure that on the left face E and S are parallel. Therefore, the outward flux is

$\phi _L =E. \Delta S= -200 \hat i \Delta S$

$\qquad =+200 \Delta S$ since $\hat i . \Delta S =- \Delta S$

$\qquad= +200 \times \pi (0.05)^2 =+ 1.57 N m^2 C^{-1}$

On the right face, E and $\Delta S$ are parallel and therefore

$\phi _R= E. \Delta S =+ 1.57 Nm^2 C^{-1}$

Hence A is the correct answer.

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