Answer
$\rho = 4400~kg/m^3$
Work Step by Step
We can find the mass $M$ of the asteroid:
$T^2 = \frac{4\pi^2~r^3}{GM}$
$M = \frac{4\pi^2~r^3}{GT^2}$
$M = \frac{(4\pi^2)(1.0\times 10^5~m)^3}{(6.67\times 10^{-11}~N~m^2/kg^2)(27\times 3600~s)^2}$
$M = 6.3\times 10^{16}~kg$
We can find the density of the asteroid:
$\rho = \frac{M}{V}$
$\rho = \frac{6.3\times 10^{16}~kg}{14,100\times 10^9~m^3}$
$\rho = 4400~kg/m^3$
