Answer
At a distance of $~~\sqrt{3}~R~~$ from the center of the planet, the gravitational acceleration is $\frac{a_g}{3}$
Work Step by Step
$a_g = \frac{GM}{R^2}$
We can find $r$ such that the gravitational acceleration is $\frac{a_g}{3}$:
$\frac{GM}{r^2} = \frac{a_g}{3}$
$\frac{GM}{r^2} = \frac{1}{3}\cdot \frac{GM}{R^2}$
$r^2 = 3R^2$
$r = \sqrt{3}~R$
At a distance of $~~\sqrt{3}~R~~$ from the center of the planet, the gravitational acceleration is $\frac{a_g}{3}$
