This is a surprisingly simple thing to calculate.

It is a well known result that a consequence of the inverse square law is that there is no force inside a symmetrical hollow shell. This means that as the object falls into the hole, it will appear to be attracted by a sphere of decreasing radius - the mass outside "doesn't count."

The acceleration of gravity at the surface of a sphere of radius R (assuming uniform density $\rho$) is given by

$$\begin{align}
a &= \frac{GM}{R^2} \\
&= \frac{4G\pi R^3\rho}{3R^2} \\
&= \frac43\pi \rho G R\\
\end{align}$$

Where $G$ is the gravitational constant, and $R$ is the distance to the center of the earth. In other words - the acceleration is proportional to the distance to the center. The corollary is that an object dropped into a hole through the center of the earth will exhibit simple harmonic motion.

Do you think you can now solve the problem yourself?