Answer
$A.)~0.301c$
Work Step by Step
We know that
$K.E=m_{\circ}c^2(\frac{1}{\sqrt{1-v^2/c^2}}-1)=eV$
This simplifies to:
$v=c\sqrt{1-(1+\frac{eV}{m_{\circ}c^2})^{-2}}$
We plug in the known values to obtain:
$v=c\sqrt{1-(1+\frac{25.0KeV}{512.24KeV})^{-2}}=0.301c$
