1. Use Stokes' law at terminal velocity

For a small spherical ball falling through a viscous medium, the viscous drag is $F_d = 6 \pi \eta r v_t$ where $r$ is the radius and $v_t$ is the terminal velocity.

Since the density of the medium is negligible, buoyant force can be ignored. Thus at terminal velocity, $\text{weight} = \text{viscous drag}$

So, $mg = 6\pi \eta r v_t$

Hence, $v_t = \frac{mg}{6\pi \eta r}$

2. Observe the dependence on radius

In this question, the mass remains the same and the medium is the same, so $m$ and $\eta$ are constant. Thus, $v_t \propto \frac{1}{r}$

3. Compare the two balls

For the first ball of radius $r$, $v_1 = v$

For the second ball of radius $2r$, $v_2 = \frac{1}{2} v_1 = \frac{v}{2}$

4. Match with the options

Thus the terminal velocity of the second ball is $\boxed{\frac{v}{2}}$

So the correct option is D.