Answer
$\Delta V = 3.1\times 10^{6}~V$
Work Step by Step
We can find $\gamma$:
$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
$\gamma = \frac{1}{\sqrt{1-\frac{(0.99~c)^2}{c^2}}}$
$\gamma = \frac{1}{\sqrt{1-0.9801}}$
$\gamma = 7.09$
We can find the kinetic energy after the electron has been accelerated:
$K = (\gamma -1)~mc^2$
$K = (7.09 -1) (9.109\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2$
$K = 5.0\times 10^{-13}~J$
We can find the required potential difference:
$\Delta V~q = K$
$\Delta V = \frac{K}{q}$
$\Delta V = \frac{5.0\times 10^{-13}~J}{1.6\times 10^{-19}~C}$
$\Delta V = 3.1\times 10^{6}~V$