Answer
$1.2\times10^{5}V $.
Work Step by Step
Calculate $V_A-V_B$. Let 0.1 m = L.
$$ V_A-V_B =(\frac{kq_1}{L}+\frac{kq_2}{\sqrt{2}L})-(\frac{kq_2}{L}+\frac{kq_1}{\sqrt{2}L})$$
$$=\frac{k}{L}(q_1-q_2) (1-\frac{1}{\sqrt{2}})$$
$$=\frac{8.99\times10^9}{0.1m}(4.5\times10^{-6}C) (1-\frac{1}{\sqrt{2}})=1.2\times10^{5}V $$
