= Solution
In SI units <Maxwell's equations> are
$$
\nabla\cdot E=\rho_e/\epsilon_0,\quad\nabla\cdot B=0,\quad\nabla\times E=-\partial_tB,\quad\nabla\times B=\mu_0J+\mu_0\epsilon_0\partial_tE.
$$
The given vector <identity element>[identity] yields
$$
\nabla\cdot(E\times B)=B\cdot(\nabla\times E)-E\cdot(\nabla\times B)=-B\cdot\partial_tB-\mu_0E\cdot J-\mu_0\epsilon_0E\cdot\partial_tE.
$$
Divide by $\mu_0$ and collect the two energy <derivatives> to obtain the local <identity element>[identity] $\partial_t\mathcal U+\nabla\cdot P+E\cdot J=0$. Integration over the fixed region and the <divergence theorem> give the required integral conservation law. Its terms are, respectively, the rate of change of stored <electromagnetic energy>, the outward <electromagnetic energy> flux through the boundary, and work per unit time done by the <field> on charges. The last term transfers energy from the <field> to matter when positive.
In vacuum put $\rho_e=0$, $J=0$ and write $\phi=kz-\omega t$. <Faraday's law> gives the plane-wave <magnetic field>
$$
\boxed{B=(0,\alpha k\cos\phi/\omega,0).}
$$
Both divergence equations are satisfied, and the remaining curl equation gives $k^2=\mu_0\epsilon_0\omega^2$. Hence $\boxed{\omega=ck,\quad c=(\mu_0\epsilon_0)^{-1/2}}$, with $B_y=E_x/c$. Here the corresponding <field> is the pure wave, with no added static background.
Its electric and magnetic energy densities are equal. Explicitly,
$$
\boxed{\mathcal U=\epsilon_0\alpha^2\cos^2\phi,\qquad P=\epsilon_0c\alpha^2\cos^2\phi\,\hat z,\qquad|P|=c\mathcal U.}
$$
The energy flux points in the propagation direction, toward increasing $z$.
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