Two identical charged spheres are suspended by strings of equal...

In air:

 $tan\dfrac{\theta}{2} = \dfrac{F}{mg} = \dfrac{{q}^{2}}{4\pi{ε}_{0}{r}^{2}mg}$

In water:

 $tan\dfrac{\theta}{2} = \dfrac{{F}^{'}}{m{g}^{'}} = \dfrac{{q}^{2}}{4\pi{ε}_{0}{ε}_{r}{r}^{2}{mg}_{eff}}$

Equate both equations, we get

${ε}_{0}g = {ε}_{0}{ε}_{r}g\left[1 - \dfrac{1}{1 . 5}\right]$

$⇒ {ε}_{r} = 3$