Answer
At a distance of $~~\frac{R}{2}~~$ below the surface of the planet, the magnitude of the gravitational force on the apple is $~~\frac{1}{2}~F_R$
Work Step by Step
At the surface of the planet:
$a_g = \frac{GM}{R^2}$
We can write an expression for $F_R$:
$F_R = m~a_g$
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 $r \leq R$.
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 an expression for the mass of the sphere with a radius $r$:
$M_r = \frac{M~r^3}{R^3}$
We can find $r$ such that the gravitational acceleration is $\frac{a_g}{2}$:
$\frac{GM_r}{r^2} = \frac{a_g}{2}$
$\frac{G~ (\frac{M~r^3}{R^3})}{r^2} = \frac{1}{2}\cdot \frac{GM}{R^2}$
$\frac{G~ M~r^3}{R^3~r^2} = \frac{1}{2}\cdot \frac{GM}{R^2}$
$r = \frac{R}{2}$
At a distance of $~~\frac{R}{2}~~$ from the center of the sphere, the gravitational acceleration is $\frac{a_g}{2}$
The distance from the surface of the planet is $(R-\frac{R}{2})$ which is $~~\frac{R}{2}$
At a distance of $~~\frac{R}{2}~~$ below the surface of the planet, the magnitude of the gravitational force on the apple is $~~\frac{1}{2}~F_R$
