When the ring is in perpendicular position, magnetic flux linked with it is maximum.
i.e., $\phi_1$ =BA = $B \times \pi r^2 = 2 \times \pi \times 10^{-2} = 2\pi \times 10^{-2}$ Wb (r = 10 cm = 0.1 m)
When ring is in parallel position, magnetic flux linked with it is zero.
i.e., $\phi_2$ = 0
$\therefore$ Change in magnetic flux, $\Delta \phi = \phi_1 - \phi_2 = 2\pi \times 10^{-2}$ Wb
Given, R = 0.01 $\Omega = 10^{-2} \Omega$
Using, charge, $q = \frac{\Delta \phi}{R} $, we have, $q = \frac{2 \pi \times 10^{-2}}{10^{-2}} = 2\pi $ C