Answer
The required work is $~~1.1~MeV$
Work Step by Step
We can find $\gamma_i$:
$\gamma_i = \frac{1}{\sqrt{1-\beta_i^2}}$
$\gamma_i = \frac{1}{\sqrt{1-0.98^2}}$
$\gamma_i = 5.0252$
We can find an expression for the initial kinetic energy of the electron:
$K_i = mc^2(5.0252-1)$
$K_i = 4.0252~mc^2$
We can find $\gamma_f$:
$\gamma_f = \frac{1}{\sqrt{1-\beta_f^2}}$
$\gamma_f = \frac{1}{\sqrt{1-0.19^2}}$
$\gamma_f = 7.0888$
We can find an expression for the final kinetic energy of the electron:
$K_f = mc^2(7.0888-1)$
$K_f = 6.0888~mc^2$
We can find the required work:
$Work = \Delta K$
$Work = K_f-K_i$
$Work = (6.0888~mc^2)-(4.0252~mc^2)$
$Work = 2.0636~mc^2$
$Work = (2.0636)~(511\times 10^3~eV)$
$Work = 1.1\times 10^6~eV$
$Work = 1.1~MeV$
The required work is $~~1.1~MeV$
