The orbital speed ($v_o$) of a satellite revolving in a circular orbit of radius $r$ around the Earth is given by $v_o = \sqrt{\frac{GM}{r}}$, where $G$ is the gravitational constant and $M$ is the mass of the Earth. From this relation, velocity is inversely proportional to the square root of the orbital radius: $v \propto \frac{1}{\sqrt{r}}$. Therefore, the ratio of speeds is $\frac{v_B}{v_A} = \sqrt{\frac{r_A}{r_B}}$. Given: $r_A = 4R$, $r_B = R$, and $v_A = 3v$. Substituting these values: $\frac{v_B}{3v} = \sqrt{\frac{4R}{R}} = \sqrt{4} = 2$. $v_B = 2 \times 3v = 6v$.