Answer
1.08 mWb
Work Step by Step
Magnetic flux along the flat top face:
$\Phi_{1}=BA=0.30\,T\times\pi(2.0\times10^{-2}\,m)^{2}$
$=0.38\,mWb$
Magnetic flux along the flat bottom face:
$\Phi_{2}=0.70\,mWb$ (given)
Since both $\Phi_{1}$ and $\Phi_{2}$ are directed outwards, both are taken as positive.
Magnetic flux through the curved part of the surface:
Let it be denoted by $\Phi_{curved}$. Then
According to Gauss's law, $\Phi_{1}+\Phi_{2}+\Phi_{curved}=0$
$\implies \Phi_{curved}=0-\Phi_{1}-\Phi_{2}$
$=0-0.38\,mWb-0.70\,mWb=-1.08\,mWb$
Magnitude= 1.08 mWb
