Answer
$v = 150 \mathrm{~m/s}$
Work Step by Step
From the conservation law, we could get the speed
\begin{gather}
K_{a}+U_{a} =K_{b}+U_{b} \\
K_{a} =U_{b} \\
\frac{1}{2} m v^{2} =\frac{1}{4 \pi \epsilon_{0}} \frac{Q q}{r}\tag{Solve for $v$}\\
v =\sqrt{\frac{1}{4 \pi \epsilon_{0}}\left[\frac{2 Q q}{m r}\right]} \\
v = \sqrt{\left(9.0 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)\left[\frac{2\left(5.00 \times 10^{-6} \mathrm{C} \times 3.00 \times 10^{-6} \mathrm{C}\right)}{\left(6.0 \times 10^{-5} \mathrm{kg}\right)(0.20 \mathrm{m})}\right]} \\
v = \boxed{150 \mathrm{~m/s}}
\end{gather}