Answer
$10\;\rm nC$
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 $q_1=q$, $q_2=-5$ nC which is the upper to the right one, and $q_3=5$ nC which is the lower to the left one.
$$V_{net}=V_{\rm 4nC}+V_{\rm 2nC,right}+V_{\rm 2nC,left}$$
$$V_{net}=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}+\dfrac{kq_3}{r_3}$$
Solving for $q_1$,
$$\dfrac{kq_1}{r_1} =V_{net}-\dfrac{kq_2}{r_2}-\dfrac{kq_3}{r_3}$$
$$ q_1 =r_1\left[\dfrac{V_{net}}{k}-\dfrac{ q_2}{r_2}-\dfrac{ q_3}{r_3}\right]$$
Plug the known from the given figure,
$$q =\left(\sqrt{0.02^2+0.04^2}\right)\left[\dfrac{3140}{9\times 10^9}-\dfrac{ -5\times 10^{-9}}{0.04}-\dfrac{ 5\times 10^{-9}}{0.02}\right]$$
$$q=\color{red}{\bf 10}\;\rm nC$$
