Through what potential difference must an electron be accelerated...

The rest mass energy of an electron is $m_0 c^2 = 0.511 \text{ MeV}$.

The kinetic energy of a relativistic particle is given by $K = (\gamma - 1)m_0 c^2$, where $\gamma = \dfrac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}}$.

Given $v = 0.99c$, we calculate $\gamma$ as:

$\gamma = \dfrac{1}{\sqrt{1 - (0.99)^2}} = \dfrac{1}{\sqrt{1 - 0.9801}} = \dfrac{1}{\sqrt{0.0199}} \approx 7.089$

The kinetic energy is:

$K = (7.089 - 1) \times 0.511 \text{ MeV} = 6.089 \times 0.511 \text{ MeV} \approx 3.11 \text{ MeV}$

Since the kinetic energy is gained by accelerating the electron through a potential difference $V$, we have $K = eV$.

$eV = 3.11 \text{ MeV} \Rightarrow V = 3.1 \text{ MV}$