Answer
$T = \sqrt{\frac{3\pi}{G~\rho}}$
Work Step by Step
We can find an expression for the mass $M$:
$\rho = \frac{M}{V}$
$M = V~\rho$
$M = \frac{4}{3}\pi~R^3~\rho$
Let $m$ be the mass of some material at the equator. Suppose that the gravitational force is equal in magnitude to the centripetal force. We can find an expression for the speed $v$ of a point at the equator:
$\frac{GMm}{R^2} = \frac{mv^2}{R}$
$v^2 = \frac{GM}{R}$
$v^2 = \frac{G(\frac{4}{3}\pi~R^3~\rho)}{R}$
$v^2 = \frac{4G\pi~R^2~\rho}{3}$
$v = \sqrt{\frac{4G\pi~R^2~\rho}{3}}$
We can find the period:
$T = \frac{distance}{speed}$
$T = \frac{2\pi~R}{\sqrt{\frac{4G\pi~R^2~\rho}{3}}}$
$T = \frac{\sqrt{12\pi^2~R^2}}{\sqrt{4G\pi~R^2~\rho}}$
$T = \sqrt{\frac{12\pi^2~R^2}{4G\pi~R^2~\rho}}$
$T = \sqrt{\frac{3\pi}{G~\rho}}$
