The acceleration due to gravity on the earth's surface at the poles...

Acceleration due to gravity at height $h$ above the pole

is given by

${g}_{h} = g\left(1 - \dfrac{2h}{R}\right) . . . ( 1 )$

Acceleration due to gravity at the equator due to rotation of the earth is given by

$g ' = g - {\omega}^{2}R\cos^{2}\lambda . . . ( 2 )$

At the equator, $\lambda = 0$

$⇒ g ' = g - {\omega}^{2}R . . . ( 3 )$

As both weights are equal, i.e.; ${g}_{h} = {g}^{'}$

From equation $( 1 )$ and $( 2 )$, we get,
$⇒ {\omega}^{2}R = \dfrac{2g}{R}h$
$∴ h = \dfrac{{R}^{2}{\omega}^{2}}{2g}$