Answer
$1.01\times 10^5\;\rm m/s$
Work Step by Step
It is obvious, from the given graph, that the proton moves into a lower potential region, so its
speed will increase.
And since the energy is conserved,
$$E_i=E_f$$
$$E_i+U_{i}=K_ f+U_{f}$$
$$\frac{1}{2}m v_i^2+(eV_i)=\frac{1}{2}mv_f^2 +(eV_f)$$
$$\frac{1}{2}m v_i^2+(eV_i)-(eV_f)=\frac{1}{2}mv_f^2 $$
where $i\rightarrow$ is at point A, and $f\rightarrow$ is at point B.
$$ v_A^2+\dfrac{(2eV_A)-(2eV_B)}{m}= v_B^2 $$
Hence,
$$ v_B=\sqrt{ v_i^2+\dfrac{ 2e[V_A -V_B]}{m} }$$
Plug the known;
$$ v_B=\sqrt{ (50,000)^2+\dfrac{ 2(1.6\times 10^{-19})[30-(-10)]}{1.67\times 10^{-27}} }$$
$$v_B=\color{red}{\bf 100,820}\;\rm m/s$$
