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

The speed of the electron is $v = 0.6c$, which is comparable to the speed of light, so relativistic mechanics must be used.

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

Substituting $v = 0.6c$:

$\gamma = \dfrac{1}{\sqrt{1 - (0.6)^2}} = \dfrac{1}{\sqrt{1 - 0.36}} = \dfrac{1}{\sqrt{0.64}} = \dfrac{1}{0.8} = 1.25$

The kinetic energy gained by the electron is:

$K = (1.25 - 1)m_0 c^2 = 0.25 m_0 c^2$

The rest mass energy of an electron is $m_0 c^2 \approx 511 \text{ keV}$.

$K = 0.25 \times 511 \text{ keV} = 127.75 \text{ keV} \approx 128 \text{ keV}$

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

$eV = 128 \text{ keV} \Rightarrow V = 128 \text{ kV}$