A planet is a uniform sphere of radius R = 3200 km and centre O…

Step 1. Superposition. The planet with the cavity is the full planet (density $\rho$) plus a sphere of density $-\rho$ filling the cavity. Inside a uniform sphere, the field at a point with position $\vec r$ from its centre is $-\tfrac43\pi G\rho\,\vec r$. For a point P in the cavity:

This is the same vector at every point of the cavity: the field there is uniform, pointing from C towards O. Before the digging the surface field was $g_0 = \tfrac43\pi G\rho R = 4.0\ \text{m/s}^2$, and $OC = R/2$, so the field in the cavity is $\dfrac{g_0}{2} = 2.0\ \text{m/s}^2$.

Step 2. The fall through the cavity. S, C and O lie on one line, along the field. So the stone falls straight from S to O, a distance $R$, with a constant acceleration of $2.0\ \text{m/s}^2$:

Step 3. The full planet. Inside it the field is $g_0\dfrac{r}{R}$, towards the centre, so the stone moves in simple harmonic motion with $\omega = \sqrt{\dfrac{g_0}{R}}$. From rest at the surface to the centre is a quarter of a period: $t = \dfrac{\pi}{2}\sqrt{\dfrac{R}{g_0}} = \dfrac{\pi}{2} \times 894.4\ \text{s} = 1405\ \text{s}$. The speed at the centre is $\omega R = \sqrt{g_0R} = 3578\ \text{m/s}$.

The two speeds are equal. On the way down, the work done per kilogram is the area under the graph of field against distance: a triangle of height $g_0$ and base $R$ for the full planet, and a rectangle of height $g_0/2$ and the same base for the cavity. Both areas are $\tfrac12 g_0R$. The stone in the full planet arrives sooner because it feels a strong pull at the start.

Step 4. The point S'. The full planet pulls with $g_0 = 4.0\ \text{m/s}^2$ towards O. The removed sphere had $\tfrac18$ of the mass, centred at C, a distance $\tfrac32 R$ from S'. It would have pulled with $\dfrac{G(M/8)}{(1.5R)^2} = \dfrac{g_0}{18} = 0.222\ \text{m/s}^2$, also towards O. Without it: $4.0 - 0.222 = 3.78\ \text{m/s}^2$, towards O.