(c) When terminal velocity $v$ is reaching, then
$F = 2 \times 10^{- 5}v = \frac{4}{3}\pi\ r^{3}\rho g$
$2 \times 10^{- 5}v = \frac{4}{3} \times \frac{22}{7} \times \left( 1.5 \times 10^{- 3} \right)^{3} \times 10^{3} \times 10$
On solving, $v = 7.07ms^{- 1} \approx 7ms^{- 1}$