Answer
$a_g = 9.79~m/s^2$
Work Step by Step
We can find the volume of a sphere at a distance $r$ from the center of the sphere.
$V_r = \frac{4}{3}\pi~r^3$
We can assume that $r \leq R$, where $R$ is the radius of the Earth.
We can express $V_r$ as a fraction of the original volume $V_R$:
$\frac{V_r}{V_R} = \frac{\frac{4}{3}\pi~r^3}{\frac{4}{3}\pi~R^3} = \frac{r^3}{R^3}$
We can find the mass of the part of the Earth with a radius of $6345~km$:
$M_r = \frac{M~r^3}{R^3}$
$M_r = \frac{(5.98\times 10^{24}~kg)~(6.345\times 10^6~m)^3}{(6.37\times 10^6~m)^3}$
$M_r = 5.91\times 10^{24}~kg$
We can find $a_g$ at a depth of $25.0~km$:
$a_g = \frac{GM_r}{r^2}$
$a_g = \frac{(6.67\times 10^{-11}~N~m^2/kg^2)(5.91\times 10^{24}~kg)}{(6.345\times 10^6~m)^2}$
$a_g = 9.79~m/s^2$
