Answer
$v_B = 7.41 \,\text{m/s}$
Work Step by Step
We will use the conservation energy law to get the speed
\begin{gather}
K_{A}+U_{A}= K_{B}+U_{B} \\
\frac{1}{2} m v_{A}^{2}+ qV_{A} =\frac{1}{2} m v_{B}^{2}+qV_{B} \\
\frac{1}{2} m v_{B}^{2} =\frac{1}{2} m v_{A}^{2}+q( V_{A}-V_{B})\\
\frac{1}{2} m v_{B}^{2} = \frac{1}{2} (2.00 \times 10^{-4} \mathrm{kg}) (5.00 \mathrm{~m/s})^{2}+5 \times 10^{-6} \,\text{C}( +200 \mathrm{V} -800 \mathrm{V})\\
\frac{1}{2} m v_{B}^{2} = 0.0055 \,\text{J}
\end{gather}
The speed will be
\begin{align}
v_B &= \sqrt{\dfrac{2 (0.0055 \,\text{J})}{m}}\\
&= \sqrt{\dfrac{2 (0.0055 \,\text{J})}{2.00 \times 10^{-4} \mathrm{kg}}}\\
&= \boxed{7.41 \,\text{m/s}}
\end{align}