Answer:
Option B
Explanation:
The electric field intensity due to the point charge q placed at the centre of a sphere of radius r,
$E=\frac{1}{4\pi \epsilon_{0}}\frac{q}{r^{2}}$
As the balloon is blown up, the charge enclosed by the Gaussian surface remains the same. So, electric field intensity remains the same. The electric flux is given by
$\phi= E.dS$
$\Rightarrow$ $ \phi\propto E$
As E remains the same, the flux remains unchanged
