Answer
Student 1 is correct .
Work Step by Step
As the electric flux passing through a surface depends on the charge enclosed by the surface.
That is, $\phi=E.dA$,
where $E$ is electric field and $dA$ is surface area.
So, flux depends on both electric field and surface area.
In case of surface A its surface area is small but its electric field is large while in case of B its surface area is large but its electric field is small. Thus,
$\phi=\frac{q}{\epsilon_{0}}$
and the charge is same for both surface A and surface B, therefore, the flux passing through them will also be same.
Hence, Student 1 is correct.