Answer
At a distance of $~~0.414~R~~$ above 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 $r$ such that the gravitational acceleration is $\frac{a_g}{2}$:
$\frac{GM}{r^2} = \frac{a_g}{2}$
$\frac{GM}{r^2} = \frac{1}{2}\cdot \frac{GM}{R^2}$
$r^2 = 2R^2$
$r = \sqrt{2}~R$
$r = 1.414~R$
At a distance of $~~1.414~R~~$ from the center of the planet, the gravitational acceleration is $\frac{a_g}{2}$
The distance from the surface of the planet is $(1.414~R-R)$ which is $~~0.414~R$
At a distance of $~~0.414~R~~$ above the surface of the planet, the magnitude of the gravitational force on the apple is $~~\frac{1}{2}~F_R$
