Answer
$\approx -1.6\;\rm kV$
Work Step by Step
We know that the net potential is the sum of the potentials due to each charge at the needed dot.
Let's assume that $ q_1=1$ nC is the upper one, $ q_2=-2$ nC is the left one, and $q_3=-2$ nC is the right one.
$$V_{net}=V_{\rm 4nC}+V_{\rm -2nC,right}+V_{\rm -2nC,left}$$
$$V_{net}=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}+\dfrac{kq_3}{r_3}$$
$$V_{net}=k\left[\dfrac{ q_1}{r_1}+\dfrac{ q_2}{r_2}+\dfrac{ q_3}{r_3}\right]$$
It is obvious that the 3 charges form an equilateral triangle, and that the given dot is just at the center of this triangle. Hence $r_1=r_2=r_3=r$.
$$V_{net}=k\left[\dfrac{ q_1}{r}+\dfrac{ q_2}{r}+\dfrac{ q_3}{r}\right]$$
$$V_{net}=\dfrac{k}{r}\left[q_1+ q_2+ q_3 \right]\tag 1$$
From the geometry of the right triangle, in the figure below, we can see that $\cos30^\circ=\dfrac{1.5}{r_1}$, and hence,
$$r_1=\dfrac{1.5}{\cos30^\circ}=\bf \sqrt 3\;\rm cm$$
Plug the known into (1)
$$V_{net}=\dfrac{(9\times 10^9)}{\sqrt{3}\times 10^{-2}}\left[1-2-2\right]\times 10^{-9}$$
$$V_{net}=\color{red}{\bf -1559}\;\rm V$$
