Old tunnel problem

According to Newton's law of gravitation:

$F(r) = -G \frac{m M(r)}{r^2}$

is the force a body of mass $m$ feels when attracted to the Earth due its gravity when the distance between $m$ and Earth's center is $r$ and $M(r)$ is the mass of the Earth inside a sphere of radius $r$ and same center as the Earth. So

$M(r) = \rho V(r) = \frac{4}{3} \pi \rho r^{3},$

therefore

$F(r) = - \frac{4}{3} \pi G \rho m r.$

Now, according to Newton's second law of motion in this case we may write $F(r) = m r''$ . So

$r'' + \frac{4}{3} \pi G \rho r = 0,$

and due to our knowledge of SHM we may write

$\omega^{2} = \left( \frac{2 \pi}{T} \right)^{2} = \frac{4}{3} \pi G \rho \Rightarrow T = \sqrt{ \frac{3 \pi}{G \rho} }.$

But, since we asked for the time the object takes to reach the other side:

$t = \frac{1}{2} \left( \frac{3 \pi}{G \rho} \right)^{1/2}.$

So $A = 2$ , $B = 3$ and $C = 2$ .