Answer
$47.4\,\mu Wb$
Work Step by Step
$\Phi_{1}=-25.0\,\mu Wb$ (direction is inwards. Therefore the sign is negative)
$\Phi_{2}=B\cdot A=BA\cos 0^{\circ}$
$=1.60\times10^{-3}\,T\times\pi(12.0\times10^{-2}m)^{2}$
$=7.24\times10^{-5}\,Wb=72.4\,\mu Wb$
Let us denote the net magnetic flux through the curved surface by $\Phi_{3}$.
According to Gauss's law, $\Phi_{1}+\Phi_{2}+\Phi_{3}=0$
Or $\Phi_{3}=0-\Phi_{1}-\Phi_{2}=0-(-25.0\,\mu Wb)-72.4\,\mu Wb$
$= -47.4\,\mu Wb$
Magnitude= $47.4\,\mu Wb$
