$$
F(r)=k T
$$
now $F_{\text {net }}$ on a particle is $2 F_{q} \cos 30^{\circ}$ due to the other two charges
$$
\mathrm{F}_{\mathrm{net}}=\frac{2 \mathrm{kq}^{2}}{\mathrm{a}^{2}} \times \frac{\sqrt{3}}{2}
$$
also $r=\frac{2}{3}\left(\frac{\sqrt{3}}{2} a\right)$
$\therefore \mathrm{a}=\sqrt{3} \mathrm{r}$ replacing it in $\mathrm{F}_{\text {nat }}$ we get
$$
\mathrm{F}_{\mathrm{net}}=\frac{2 \mathrm{kq}^{2}}{(\sqrt{3} \mathrm{r})^{2}} \times\left(\frac{\sqrt{3}}{2}\right)=\frac{\mathrm{kq}^{2}}{\sqrt{3} \mathrm{r}^{2}}
$$
this is balanced by $\mathrm{F}(\mathrm{r})$
$$
\begin{array}{l}
\therefore F(r)=F_{n e t} \Rightarrow k r=\frac{1 \times q^{2}}{4 \pi \varepsilon_{0} \times \sqrt{3} r^{2}} \\
\therefore r=\left(\frac{\sqrt{3} q^{2}}{12 \pi \varepsilon_{0} k}\right)^{1 / 3}
\end{array}
$$