Time and tide waits for no man!

One can try adding all the contributions from all the charges, to find the resultant electric field. However, there is a nice trick that one can use to find the same field. Their are two main key ideas:

(1) Electric fields satisfy the superposition principle and hence it does not matter how we combine them to get the field. Hence we can use a way that would make it easier for use to add them.

(2) Observe that the charges that are exactly opposite to each other (like $6$ and $12$ or $2$ and $8$ ) have the difference of $6$ and there are $6$ such pairs. Also, since the charges are all negative, the resultant field at the center due to two oppositely placed charges is directed towards the larger of the two.

It is easy to see that we get $6$ resultant vectors each of magnitude $6$ (assuming that distance between each charge from the pair is $1$ ) and angle between two adjacent vectors is $\frac{2\pi}{12} = \frac{\pi}{6}$ . These vectors are aligned towards the numbers $7, 8, 9, 10, 11$ and $12$ . The vertical contributions of vectors $7$ and $11$ cancel each other. Their horizontal contributions add up to : $2 \times 6 \times \cos(\frac{\pi}{3})$ . Similarly, the vertical contributions of vectors $8$ and $10$ add up and horizontal contributions sum up to : $2\times 6 \times \cos(\frac{\pi}{6})$ . Adding the contribution $6$ by the completely horizontal vector in the direction of $9$ , we get the total horizontal component of the resultant vector to be:

$E_{h} = 2 \times 6 \times \frac{1}{2} + 2 \times 6 \times \frac{\sqrt{3}}{2} + 6 = 6(2+\sqrt{3})$

The net vertical contribution is provided only by the vector pointing in the direction of $12$ and its magnitude is $6$ .

$E_{v} = 6$

Thus, the resultant field makes an angle of $\theta = \tan^{-1}(\frac{E_{v}}{E_{h}}) = \frac{\pi}{12}$ with the vector in the direction of $9$ . Thus, the resultant time must be $\boxed{9:30}$