Answer
The acceleration due to gravity will be $\frac{1}{10}$ of the value at the Earth's surface when the distance from the Earth's center is 20,200 km.
Work Step by Step
$\frac{1}{10}\times G\frac{m}{r^2} = G\frac{m}{(\sqrt{10}~r)^2} $
We can see that the acceleration due to gravity will be $\frac{1}{10}$ of the value at the Earth's surface when the distance is $\sqrt{10} ~r$.
$\sqrt{10}~r = \sqrt{10}\times(6380 ~km) = 20,200 ~km$
The acceleration due to gravity will be $\frac{1}{10}$ of the value at the Earth's surface when the distance from the Earth's center is 20,200 km.
