$T\sin \theta =F=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{a^2}\mathrm{................}\left(i\right)$

$T\cos \theta =mg\mathrm{...............}\left(ii\right)$

From (i) and (ii), $\tan \theta =\frac{q^2}{4\pi {\epsilon }_0{a}^{2}mg}$

or $\frac{a/2}{L}=\frac{q^2}{4\pi {\epsilon }_0{a}^{2}mg}(\sin ceforsmallq,\tan \theta \approx a/2L)$

or $\frac{a^3}{L}=\frac{q^2}{2\pi {\epsilon }_{0}mg}\mathrm{.............}\left(iii\right)$

When one of the balls is discharged, the balls come closer and touch each other and again separate due to repulsion. The charge on each ball after touching each other is q/2. Replacing q with q with q/2 in Eq. (iii), we get

$\frac{b^3}{L}=\frac{{(q/2)}^2}{2\pi {\epsilon }_{0}mg}\mathrm{.............}\left(iv\right)$

$From Eqs. \left(iii\right) and \left(iv\right),$

$\frac{b^3}{a^3}=\frac{1}{4}or\frac{a^3}{b^3}=4$