A clock dial has point charges $-\mathrm{q},-2 \mathrm{q},-3 \mathrm{q}, \ldots \ldots \ldots .-12 \mathrm{q}$ at the positions of the corresponding numbers on the dial respectively. The time at which the hour's hand points the direction of the net electric field at the centre of the dial is (Assume clock hands do not influence the net electric (field)

Electric field is given by $\mathrm{E}=\frac{\mathrm{Kq}}{\mathrm{r}^2}$ Charge is more then electric field is also more. $\therefore \mathrm{E}_1=\mathrm{E}_2=\mathrm{E}_3=\mathrm{E}_4=\mathrm{E}_5=\mathrm{E}_6=\mathrm{E}$
The magnitude of two vector and their resuiant is same then the angle between them is $120^{\circ}$. Starting from $\mathrm{E}_1$, the other vector for which the difference in angle is $120^{\circ}$ with $\mathrm{E}_1$ would be $\mathrm{E}_5$. The resultant of these vectors would be in the direction of $\mathrm{E}_3$ having magnitude is $2 \mathrm{E}$. Similarly for $\mathrm{E}_2$. Since, $E_3$ and $E_4$ represents $-10 q$ and $-9 q$ on the clock, their bisector would indicate a time of 9:30.