Concept:
For a satellite in circular orbit:
\[
v=\sqrt{\frac{GM}{r}}
\]
So,
\[
v\propto \frac{1}{\sqrt{r}}
\]
ip
Step 1: Use the ratio of orbital speeds.
For satellite \(P\):
\[
r_P=3R,\qquad v_P=2V
\]
For satellite \(Q\):
\[
r_Q=R
\]
Now,
\[
\frac{v_Q}{v_P}=\sqrt{\frac{r_P}{r_Q}}
=\sqrt{\frac{3R}{R}}=\sqrt{3}
\]
ip
Step 2: Find \(v_Q\).
\[
v_Q=\sqrt{3}\,v_P
=\sqrt{3}(2V)=2\sqrt{3}V
\]
ip
Hence, the correct answer is:
\[
\boxed{(A)\ 2\sqrt{3}V}
\]