Answer
$\approx 1.8\;\rm kV$
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 $q_1=4$ nC, $q_2=2$ nC the upper to the right one, and $q_3=2$ nC the lower to the left one.
$$V_{net}=V_{\rm 4nC}+V_{\rm 2nC,right}+V_{\rm 2nC,left}$$
$$V_{net}=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}+\dfrac{kq_3}{r_3}$$
$$V_{net}=k\left[\dfrac{ q_1}{r_1}+\dfrac{ q_2}{r_2}+\dfrac{ q_3}{r_3}\right]$$
Plug the known from the given figure,
$$V_{net}=(9\times 10^9)\left[\dfrac{ 4}{\sqrt{0.03^2+0.04^2}}+\dfrac{ 2}{0.04}+\dfrac{ 2}{0.03}\right]\times 10^{-9}$$
$$V_{net}=\color{red}{\bf 1770}\;\rm V$$