d
$6 \pi \eta r v _{ t }=\frac{4}{3} \pi r ^{3} \rho g$
$v _{ t }=\frac{4}{3} \times \frac{\pi r^{3} \rho g }{6 \pi \eta r}$
$v _{ t }=\frac{4}{3} \times \frac{\pi r ^{3} \rho g }{6 \pi \eta r}=\frac{2 \times 10^{-12} \times 10^{3} \times 10}{9 \times 1.8 \times 10^{-5}}$
$=123.4 \times 10^{-6}\,m / s$
