The mass of the spherical hailstone is $$m=\frac 43\pi r^3\rho$$ where$\rho$ is the density. So $$\frac{dm}{dt}=4\pi r^2\rho \frac{dr}{dt}=3mk$$
Meanwhile the equation of motion is $$\frac{d}{dt}(mv)=mg$$ $$\implies m\frac{dv}{dt}+v\frac{dm}{dt}=mg$$
For the limiting speed we require $\frac{dv}{dt}\rightarrow0$
In which case, $$v(3mk)\rightarrow mg$$
So the terminal velocity is $\frac{g}{3k}$