Solution

In the Earth frame, the electron takes a time $\tau$ to leave the field. Half of this time is the time it takes for the electron to reach zero velocity. In the moving frame (the frame of the field), the electron has a velocity

\begin{equation} v_1 = \frac{v - (-u)}{1 - \frac{(-u)v}{c^2}} = \frac{v + u }{1 + \frac{uv}{c^2}} \end{equation}

The time needed by the electron to reach zero velocity in the moving frame is:

\begin{equation} t_{1/2} = \frac{\tau \sqrt{1-u^2/c^2}}{2} \end{equation}

The initial momentum of the electron in the moving frame is:

\begin{equation} p = \frac{m_e v_1}{\sqrt{1-v_1^2/c^2}} \end{equation}

Applying Newton's second law, we have:

\begin{equation} \frac{\Delta p}{\Delta t} = \frac{- p}{t_{1/2}} =- e E \rightarrow E = \frac{p}{e t_{1/2}} \end{equation}

Finally:

\begin{equation} E = \frac{2 m_e ( v + u) }{e \tau (1-u^2/c^2) \sqrt{1- v^2/c^2}} \end{equation}