Answer
The required work is $~~1.0\times 10^3~eV$
Work Step by Step
We can find $\gamma_i$:
$\gamma_i = \frac{1}{\sqrt{1-\beta_i^2}}$
$\gamma_i = \frac{1}{\sqrt{1-0.18^2}}$
$\gamma_i = 1.01660$
We can find an express for the initial kinetic energy of the electron:
$K_i = mc^2(1.0166-1)$
$K_i = 0.01660~mc^2$
We can find $\gamma_f$:
$\gamma_f = \frac{1}{\sqrt{1-\beta_f^2}}$
$\gamma_f = \frac{1}{\sqrt{1-0.19^2}}$
$\gamma_f = 1.01855$
We can find an express for the final kinetic energy of the electron:
$K_f = mc^2(1.01855-1)$
$K_f = 0.01855~mc^2$
We can find the required work:
$Work = \Delta K$
$Work = K_f-K_i$
$Work = (0.01855~mc^2)-(0.01660~mc^2)$
$Work = 0.00195~mc^2$
$Work = (0.00195)~(511\times 10^3~eV)$
$Work = 1.0\times 10^3~eV$
The required work is $~~1.0\times 10^3~eV$
