a
Since density is negligible hence Buoyancy force will be negligibleAt terminal velocity.
${Mg}=6 \pi \eta \mathrm{rv}$
$\mathrm{V} \propto \frac{1}{\mathrm{r}} \quad$ (as mass is constant)
Now, $\frac{\mathrm{v}}{\mathrm{v}^{\prime}}=\frac{\mathrm{r}^{\prime}}{\mathrm{r}}$
$r^{\prime}=2 \mathrm{r}$
So, $v^{\prime}=\frac{v}{2}$