= Solution
Let $\psi=kx-\omega t$ and use bold $\mathbf B$ for the <magnetic flux density>, to distinguish it from the scalar amplitude $B$ in $E_z$. For a <plane electromagnetic wave> travelling along $+x$, <Faraday's law> gives the wave-associated field
$$
\mathbf B=\frac{k}{\omega}\hat{\mathbf x}\times\mathbf E.
$$
Hence
$$
\boxed{B_x=0,\qquad B_y=-\frac{kB}{\omega}\cos\psi,\qquad B_z=\frac{kA}{\omega}\sin\psi}.
$$
For an explicit component check, $\nabla\times\mathbf E=(0,kB\sin\psi,kA\cos\psi)$, while differentiating the displayed field gives $-\partial\mathbf B/\partial t$ equal to that vector. The field is transverse and also satisfies $\nabla\cdot\mathbf B=0$.
In vacuum, <Ampère-Maxwell equation> compatibility requires $\omega/k=c$, so each factor $k/\omega$ is $1/c$. More generally it is the inverse <phase velocity> on a suitable homogeneous-medium plane-wave branch. A time-independent background field could be added without changing the electric wave's Faraday equation; the requested propagating-wave field takes that independent background to be zero. Finally, $\mathbf E\times\mathbf B=(k/\omega)(E_y^2+E_z^2)\hat{\mathbf x}$ confirms the propagation direction and the relative signs.
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