Answer
(a) $\rho = \frac{3\pi}{GT^2}$
(b) $\rho = 5400 ~kg/m^3$
Work Step by Step
(a) The volume of a sphere is $V = \frac{4}{3}\pi r^3$
$T^2 = \frac{4\pi^2 r^3}{Gm}$
$\frac{m}{4\pi r^3} = \frac{\pi}{GT^2}$
$\frac{m}{\frac{4}{3}\pi r^3} = \frac{3\pi}{GT^2}$
$\frac{m}{V} = \frac{3\pi}{GT^2}$
This works for an orbit that is very near the surface of a planet because in that case, the orbital radius r is approximately the same as the radius of the planet.
(b) $\rho = \frac{3\pi}{GT^2} = \frac{3\pi}{(6.67 \times 10^{-11}~N\cdot m^2/kg^2)(85*60 ~s)^2}$
$\rho = 5400 ~kg/m^3$