Solution

Using the Lorentz transformation for electric and magnetic fields and the relation between these fields $E = B c$, we can calculate the electric field in
the moving frame:

\begin{equation} E_1 = \frac{E - (-v) B}{\sqrt{1-v^2/c^2}} = \frac{E - (-v) E/c}{\sqrt{1-v^2/c^2}} = E \sqrt{\frac{1+v/c}{1-v/c}} \end{equation}

The pressure on the moving wall is:

\begin{equation} \mathcal{P} = 2 \frac{I}{c} = 2 \frac{c \epsilon_0 E_1^2}{c} = 2 \epsilon_0 E_1^2 = 2 \epsilon_0 E^2 \frac{1+v/c}{1-v/c} \end{equation}

The factor of two is because there is a pressure due to the incoming flow and another due to the outgoing flow.

Also, the result in the $GHS$ system of units is:

\begin{equation} \mathcal{P} = \frac{1}{8 \pi} \epsilon_0 E^2 \frac{1+v/c}{1-v/c} \end{equation}