Answer
Same amount of electric flux passes through both the surfaces.
Work Step by Step
By Gauss Law, flux through a surface is given by:
$$\phi = \frac{q_{enc}}{\epsilon_{0}}$$
where $\phi$ is the flux through the surface
$q_{enc}$ is the net charge enclosed in the surface
For figure (a),
$$\phi_a = \frac{q_{enc}}{\epsilon_{0}} = \frac{+2\mu C}{\epsilon_{0}}$$
For figure (b),
$$\phi_b = \frac{q_{enc}}{\epsilon_{0}} = \frac{(+1-1+2)\mu C}{\epsilon_{0}} = \frac{+2\mu C}{\epsilon_{0}}$$
So through both the surfaces same amount of electric flux passes.