Answer
The required work is $~~1.27\times 10^{-14}~J$
Work Step by Step
We can find $\gamma$:
$\gamma = \frac{1}{\sqrt{1-\beta^2}}$
$\gamma = \frac{1}{\sqrt{1-0.500^2}}$
$\gamma = 1.1547$
We can find the kinetic energy of the electron at this speed:
$K = mc^2(\gamma-1)$
$K = (9.109\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2(1.1547-1)$
$K = 1.27\times 10^{-14}~J$
The required work is $~~1.27\times 10^{-14}~J$
