Solution:

Orbital speed is inversely proportional to the square root of the radius ($v \propto \frac{1}{\sqrt{r}}$). Thus, $\frac{v_B}{v_A} = \sqrt{\frac{r_A}{r_B}} = \sqrt{\frac{4R}{R}} = 2$, which means $v_B = 2 \times 3V = 6V$. Therefore, option C is correct.