= Solution
The source-free equations imply $k\cdot E_0=0$, $\omega=c|k|$, and
$$
B=\frac{k\times E_0}{\omega}\sin(k\cdot x-\omega t).
$$
Therefore
$$
S=\frac{E_0\times(k\times E_0)}{\mu_0\omega}
\sin^2(k\cdot x-\omega t)
=\epsilon_0c|E_0|^2\sin^2(k\cdot x-\omega t)\,\widehat k.
$$
It is parallel to $k$, and its period average is
$$
\langle S\rangle
=\frac12\epsilon_0c|E_0|^2\widehat k.
$$
The average electric and magnetic energy densities are equal:
$$
\langle u_E\rangle=\langle u_B\rangle
=\frac14\epsilon_0|E_0|^2.
$$
\b[Hence $\langle u\rangle=\epsilon_0|E_0|^2/2$ and $\langle S\rangle=c\langle u\rangle\widehat k$, as summarized by <energy density and flux of a plane electromagnetic wave>.]
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