Answer
$\sqrt{\frac{4\pi^2d^3}{3Gm}}$
Work Step by Step
We can find the required orbital speed as follows:
$F_{centripetal}=F_{gravity}$
$\implies 2m(\frac{v_{2m}^2}{\frac{1}{3}d})=\frac{G(2m)(m)}{d^2}$
This simplifies to:
$v^2=\frac{Gm}{3d}$.....eq(1)
We know that
$v^2=[\frac{2\pi(d/3)}{T}]^2$.....eq(2)
Now comparing eq(1) and eq(2), we obtain:
$\frac{4\pi^2d^2}{9T^2}=\frac{Gm}{3d}$
$\implies T=\sqrt{\frac{4\pi^2d^3}{3Gm}}$
