= Solution
For a wave travelling in $+x$,
$$
\mathbf B=\frac1c\widehat{\mathbf x}\times\mathbf E
=\frac{E_0}{c}(0,-\sin\omega\xi,\cos\omega\xi).
$$
At $x=0$, choose
$$
\mathbf v=\frac{qE_0}{\gamma m\omega}(0,\sin\omega t,-\cos\omega t).
$$
It is parallel to $-\mathbf B$, so the magnetic force vanishes, while its <derivative> supplies $q\mathbf E/(\gamma m)$. Constant speed then gives
$$
\gamma^2=1+\left(\frac{qE_0}{m\omega c}\right)^2.
$$
The circular radius is
$$
\boxed{R=\frac{|q|E_0}{\gamma m\omega^2}.}
$$
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