Answer
$2.53\;\rm cm/s$
Work Step by Step
We know that at the turning point, the energy is all a potential energy since the kinetic energy is zero; recall the gravitational potential energy of a ball that was fired vertically upward at the highest point on its trajectory.
In this situation, the energy is conserved.
$$E_i=E_f$$
$$U_i+K_i=U_f+K_f$$
where $_f\rightarrow$ is at a turning point at which $x=x_{\rm max}=0.08$ m.
$$U_i+K_i=qV+0$$
$$U_i+\frac{1}{2}mv_i^2=q(5000x_{\rm max}^2)$$
The maximum speed occurs when $U=0$ J,
$$0+\frac{1}{2}mv_{\rm max}^2= 5000qx_{\rm max}^2 $$
$$ mv_{\rm max}^2= 10,000qx_{\rm max}^2 $$
$$ v_{\rm max} = \sqrt{\dfrac{10,000qx_{\rm max}^2 }{m}}$$
Plugging the known;
$$ v_{\rm max} = \sqrt{\dfrac{10,000(10\times 10^{-9})(0.08)^2 }{1\times 10^{-3}}}$$
$$v_{\rm max} =0.0253\;\rm m/s=\color{red}{\bf 2.53}\;\rm cm/s$$
