Equipotential surfaces are shown in figure. Then the electric field strength will be

$$\begin{gathered} \text{ Using }\mathrm{dV}=-\overrightarrow{\mathrm{E}}·\mathrm{d}\overrightarrow{\mathrm{r}} \\ \Rightarrow \mathrm{\Delta V}=-\mathrm{E}.\mathrm{\Delta rcos\theta } \\ \Rightarrow \mathrm{E}=\frac{-\mathrm{\Delta V}}{\mathrm{\Delta rcos\theta }} \\ \Rightarrow \mathrm{E}=\frac{-(20-10)}{10\times {10}^{-2}\cos 120°} \\ =\frac{-10}{10\times {10}^{-2}\left(-\sin 30°\right)}=\frac{-{10}^2}{-1/2}=200 \mathrm{V}/\mathrm{m} \end{gathered}$$

Direction of E be perpendicular to the equipotential surface i.e., at 120° with x-axis.