Let us assume that the dipole is at the centre of the one equipotential surface which is spherical (Fig.). On an equipotential surface the net electric field strength along the tangent of it becomes zero. Thus $-E_{0} \sin \theta+E_{\theta}=0 \quad$ or $-E_{0} \sin \theta+\frac{p \sin \theta}{4 \pi \varepsilon_{0} r^{3}}=0$
Hence $\quad r=\left(\frac{p}{4 \pi \varepsilon_{0} E_{0}}\right)^{1 / 3}$
Alternate : Potential at the point, near the dipole is given by, $\varphi=\frac{\vec{p} \cdot \vec{r}}{4 \pi \varepsilon_{0} r^{3}}-\vec{E}_{0} \cdot \vec{r}+$ constant,
$=\left(\frac{p}{4 \pi \varepsilon_{0} r^{3}}-E_{0}\right) \cos \theta+$ Const
For $\varphi$ to be constant,
$\frac{p}{4 \pi \varepsilon_{0} r^{3}}-E_{0}=0$ or, $\frac{p}{4 \pi \varepsilon_{0} r^{3}}=E_{0}$
Thus $r=\left(\frac{p}{4 \pi \varepsilon_{0} E_{0}}\right)^{1 / 3}$