Answer
$v = 1.15 \times 10^{3} \mathrm{m} / \mathrm{s}$
Work Step by Step
First, we calculate the electric field from the surface charge density by
\begin{aligned}
E &=\frac{\sigma}{2 \epsilon_{0}} \\
&=\frac{2.34 \times 10^{-9} \mathrm{C} / \mathrm{m}^{2}}{2\left(8.854 \times 10^{-12} \mathrm{C}^{2} / \mathrm{N} \cdot \mathrm{m}^{2}\right)} \\
&=132 \mathrm{N} / \mathrm{C}
\end{aligned}
As the electric force equals Newton's force, we could get the acceleration of the charge by
\begin{aligned}
a &=\frac{e E}{m} \\ &=\frac{\left(1.60 \times 10^{-19} \mathrm{C}\right)(132 \mathrm{N} / \mathrm{C})}{\left(1.672 \times 10^{-27} \mathrm{kg}\right)} \\
&=1.26 \times 10^{10} \mathrm{m} / \mathrm{s}^{2}
\end{aligned}
From Newton's law, the speed in y-direction could be found by
$$v_{y} =v_{o}+a t =0+\left(1.26 \times 10^{10} \mathrm{m} / \mathrm{s}^{2}\right)\left(5.00 \times 10^{-8} \mathrm{s}\right) =630 \mathrm{m} / \mathrm{s} $$
Hence, the speed of the proton will be
$$ v = \sqrt{(970 \mathrm{m} / \mathrm{s})^{2}+(630 \mathrm{m} / \mathrm{s})^{2}} =\boxed{1.15 \times 10^{3} \mathrm{m} / \mathrm{s} }$$
