The energy ($W$) gained by an electron accelerating through a distance ($d$) under the influence of an electric field ($E$) is given by the work done: $W = Fd = (eE)d$. Given: Energy ($W$) = 2 eV Distance ($d$, mean free path) = $4 \times 10^{-8}$ m

Substituting the values (noting that 2 eV energy corresponds to the work done on charge $e$ moving through a potential difference of 2 Volts, or simply equating $eV$ units): $eEd = 2 \text{ eV}$ $E \times d = 2 \text{ V}$ $E = \frac{2 \text{ V}}{4 \times 10^{-8} \text{ m}}$ $E = 0.5 \times 10^8 \text{ V/m}$ $E = 5 \times 10^7 \text{ V/m}$