At what height $h$ above the Earth's surface does the acceleration due to gravity decrease by a factor of 4? The radius of the Earth is $R = 6400$ km.
The acceleration due to gravity is inversely proportional to the square of the distance from the centre of the Earth:
$$g = \frac{GM}{r^2} \Rightarrow \frac{g_h}{g} = \frac{R^2}{(R+h)^2} = \frac{1}{4}$$
$$\frac{R}{R+h} = \frac{1}{2} \Rightarrow R + h = 2R \Rightarrow h = R$$
$$h = 6400 \text{ km}$$
$h = R = 6400$ km