Assume there are two identical simple pendulum Clocks- 1 is placed...

The time period of a simple pendulum is given by, $T = 2\pi\sqrt{\dfrac{l}{g}}$.

Therefore, $T ∝ \dfrac{1}{\sqrt{g}}$.

The acceleration due to gravity at height $h$ from the surface of the earth is given by, ${g}^{'} = \dfrac{g}{{\left(R + h\right)}^{2}}$

Therefore,

$\dfrac{{T}_{1}}{{T}_{2}} = \sqrt{\dfrac{{g}_{2}}{{g}_{1}}} ⇒ \dfrac{4}{6} = \dfrac{R}{R + h} ⇒ h = \dfrac{R}{2} = 3200km$