Solution

When the electron enters the magnetic field, it will describe a circular path of radius $R$.
To calculate the radius we need to use Newton's second law for the radial direction.

\begin{equation} \frac{m_e v^2}{R} = \frac{m_e (\beta c)^2}{R} = F \end{equation}

The force acting on the electron is the magnetic force $F = e [\vec{v} \times \vec{B}]$; because the magnetic field is perpendicular to the velocity of the
electron we have $F = e \beta c B$. Finally:

\begin{equation} \frac{m_e (\beta c)^2}{R} = e \beta c B \rightarrow R = \frac{m_e \beta c}{e B} \end{equation}

Thus, the time the electron spends in the magnetic field is:

\begin{equation} t = \frac{2 \pi R}{\beta c} = \frac{2 \pi m_e}{e B} \end{equation}

And, taking into account the dilatation of the time we have:

\begin{equation} t_{field} = t \gamma = \frac{2 \pi m_e}{e B} \gamma \end{equation}