Answer
$v = 1.55\times 10^7~m/s$
Work Step by Step
We can find the Lorentz factor $\gamma$:
$K = (\gamma-1)~m~c^2$
$\gamma = 1+\frac{K}{m~c^2}$
$\gamma = 1+\frac{ 2.50~MeV}{1875.6~MeV}$
$\gamma = 1.001333$
We can find the speed:
$\gamma = 1.001333$
$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = 1.001333$
$\sqrt{1-\frac{v^2}{c^2}} = \frac{1}{1.001333}$
$1-\frac{v^2}{c^2} = (\frac{1}{1.001333})^2$
$v = \sqrt{1-(\frac{1}{1.001333})^2}~c$
$v = \sqrt{1-(\frac{1}{1.001333})^2}~(3.0\times 10^8~m/s)$
$v = 1.55\times 10^7~m/s$
