Answer is (2)
$\vec{E} \times \vec{B}=\vec{V}$
$(\mathrm{E} \hat{\mathrm{j}}) \times(\overrightarrow{\mathrm{B}})=\mathrm{V} \hat{\mathrm{i}}$
So, $\overrightarrow{\mathbf{B}}=\mathbf{B} \hat{\mathbf{k}}$
Direction of propagation is along $+z$ direction.