From an electron gun, electrons are emitted with negligible speed...

The kinetic energy of the electron after being accelerated by a potential difference $V$ is given by:

$\dfrac{1}{2}mv^2 = eV \Rightarrow v = \sqrt{\dfrac{2eV}{m}}$

The electrons enter the magnetic field at a small angle $\theta$ (since they are paraxial) with the $x$-axis. The velocity component parallel to the magnetic field is:

$v_{\parallel} = v \cos\theta \approx v$

The time period of one revolution in the uniform magnetic field is independent of the velocity and is given by:

$T = \dfrac{2\pi m}{eB}$

The electrons will refocus on the $x$-axis after completing one full revolution. The distance along the $x$-axis at which they refocus is the pitch of their helical path:

$x = v_{\parallel} T \approx v T$

Substituting the expressions for $v$ and $T$:

$x = \sqrt{\dfrac{2eV}{m}} \times \dfrac{2\pi m}{eB}$

$x = \sqrt{\dfrac{2eV}{m} \times \dfrac{4\pi^2 m^2}{e^2 B^2}}$

$x = \sqrt{\dfrac{8\pi^2 m V}{e B^2}}$