Answer
$v = 0.99995~c$
Work Step by Step
We can find the kinetic energy after being accelerated through the potential difference:
$K = \Delta V~q$
$K = (50\times 10^6~V)(1.6\times 10^{-19}~C)$
$K = 8.0\times 10^{-12}~J$
We can find $\gamma$:
$K = (\gamma -1)~mc^2$
$\gamma -1 = \frac{K}{mc^2}$
$\gamma = 1+\frac{K}{mc^2}$
$\gamma = 1+\frac{8.0\times 10^{-12}~J}{(9.109\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2}$
$\gamma = 1+97.58$
$\gamma = 98.58$
We can find the speed when $\gamma = 98.58$:
$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
$\sqrt{1-\frac{v^2}{c^2}} = \frac{1}{\gamma}$
$1-\frac{v^2}{c^2} = \frac{1}{\gamma^2}$
$\frac{v^2}{c^2} = 1-\frac{1}{\gamma^2}$
$v^2 = (1-\frac{1}{\gamma^2})~c^2$
$v = \sqrt{1-\frac{1}{\gamma^2}}~c$
$v = \sqrt{1-\frac{1}{(98.58)^2}}~c$
$v = \sqrt{0.999897}~c$
$v = 0.99995~c$
