Answer
$\sqrt{\frac{72\pi^2r_1^3}{Gm_1}}$
Work Step by Step
We can find the required orbital period as follows:
$\frac{(2\pi r_1/T)^2}{r_1}=\frac{Gm^2}{(r_1+2r_1)^2}$
As $m_2=\frac{1}{2}m_1$
$\implies \frac{4\pi^2 r_1}{T^2}=\frac{G(\frac{1}{2}m_1)}{(3r_1)^2}$
This simplifies to:
$T=\sqrt{\frac{72\pi^2r_1^3}{Gm_1}}$
