Answer
$7.8\times10^{-4o}$
Work Step by Step
a) $\theta=\arctan(\frac{F_M}{mg})$
$F_M=G\frac{m_1m_2}{r^2}=6.67\times10^{-11}\frac{Nm^2}{kg^2}\frac{mm_M}{(D_M)^2}$
$mg=G\frac{m_1m_2}{r^2}=6.67\times10^{-11}\frac{Nm^2}{kg^2}\frac{mm_E}{(r_E)^2}$
$\theta=\arctan\Big(\frac{mm_M}{(D_M)^2}\times\frac{(r_E)^2}{mm_E}\Big)=\arctan\Big(\frac{m_M(r_E)^2}{m_E(D_M)^2}\Big)$
b) $V=\frac{\pi r^2h}{3}=\frac{\pi(2000m)^2(4000m)}{3}=1.7\times10^{10}m^3$
$m=d\times V=(3000\frac{kg}{m^3})\times(1.7\times10^{10}m^3)$
$=5\times10^{13}kg$
c) $\arctan\Big(\frac{m_M(r_E)^2}{m_E(D_M)^2}\Big)=\arctan\Big(\frac{(5\times10^{13}kg)(6.38\times10^6m)^2}{(5.98\times10^{24}kg)(5\times10^3m)^2}\Big)=7.8\times10^{-4o}$