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

The speed of the electron is $v = 0.9c$, which is comparable to the speed of light, so relativistic mechanics must be used.

The relativistic factor $\gamma$ is given by:
$\gamma = \dfrac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}} = \dfrac{1}{\sqrt{1 - 0.81}} = \dfrac{1}{\sqrt{0.19}} \approx 2.294$

The kinetic energy $K$ of the electron is:
$K = (\gamma - 1)m_0 c^2$

Using the rest mass energy of an electron, $m_0 c^2 \approx 0.511 \text{ MeV}$:
$K = (2.294 - 1) \times 0.511 \text{ MeV} = 1.294 \times 0.511 \text{ MeV} \approx 0.661 \text{ MeV} = 661 \text{ keV}$

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

$V = \dfrac{661 \text{ keV}}{e} = 661 \text{ kV}$