Answer
Electric Flux through the Gaussian cubical surface is 0.45 Vm.
Work Step by Step
Electric flux through a Gaussian surface is given by:
$$\phi = \frac{q_{enc}}{\epsilon_{0}}$$
So for the given cubical Gaussian surface
$$\phi = \frac{+6\times 10^{-12}-2\times10^{-12}}{\epsilon_0} = \frac{+4\times 10^{-12}}{8.85\times 10^{-12}} =0.45\text{ Vm}$$
