Within a discharge tube filled with a gas, a molecule undergoes...

The acceleration of the electron in the electric field is given by:

$a = \dfrac{eE}{m}$

Assuming the electron starts from rest after being detached, the distance traveled between successive collisions is the mean free path $\lambda$. Using the equation of motion $s = ut + \dfrac{1}{2}at^2$ with $u = 0$ and $s = \lambda$:

$\lambda = \dfrac{1}{2}at^2 = \dfrac{1}{2} \left( \dfrac{eE}{m} \right) t^2$

Rearranging for the transit time $t$:

$t = \sqrt{\dfrac{2m\lambda}{eE}}$

Substituting the given values ($E = 5.0 \times 10^3 \text{ V m}^{-1}$, $\lambda = 1.0 \times 10^{-3} \text{ m}$, $m = 9.1 \times 10^{-31} \text{ kg}$, $e = 1.6 \times 10^{-19} \text{ C}$):

$t = \sqrt{\dfrac{2 \times 9.1 \times 10^{-31} \times 1.0 \times 10^{-3}}{1.6 \times 10^{-19} \times 5.0 \times 10^3}}$

$t = \sqrt{\dfrac{18.2 \times 10^{-34}}{8.0 \times 10^{-16}}}$

$t = \sqrt{2.275 \times 10^{-18}}$

$t \approx 1.5 \times 10^{-9} \text{ s} = 1.5 \text{ ns}$