As shown in the figure, a configuration of two equal point charges...

For the condition of equilibrium, the Coulomb force is equal to $mgsin\theta$. Hence,

$\dfrac{k{{q}_{0}}^{2}}{4{h}^{2}} = ( mgsin\theta ) ⇒ \dfrac{k{{q}_{0}}^{2}}{4{h}^{2}} = \left(20 \times {10}^{- 3}\right) \times 10 \times \dfrac{1}{2}$

$⇒ \dfrac{9 \times {10}^{9} \times 4 \times {10}^{- 12}}{4{h}^{2}} = 20 \times {10}^{- 3} \times 10 \times \dfrac{1}{2} ⇒ h = \left(\sqrt{\dfrac{9}{100}}\right)m = 0 . 3m = 300 \times {10}^{- 3}m$

${h}^{2} = \left(\dfrac{9}{100}\right) ⇒ h = \left(\dfrac{3}{10}\right)m = 0 .3 m$

$= 300 \times {10}^{- 3}m$