= Solution
Keep $\mathbf A=(0,Bx,0)$ and choose the <scalar potential> $\phi=-Ex$. Then
$$
H=\frac1{2m}\left[p_x^2+(p_y-qBx)^2+p_z^2\right]-qEx.
$$
For $p_z=0$, completing the square gives
$$
H=\frac{p_x^2}{2m}
+\frac{m\omega_c^2}{2}(x-x_c)^2
-\frac EBp_y-\frac{mE^2}{2B^2},
\qquad
x_c=\frac{p_y}{qB}+\frac{mE}{qB^2}.
$$
Hence the spectrum on $\mathbb R^3$ is
$$
\boxed{E_n(k)=\hbar\omega_c\left(n+\frac12\right)
-\frac EB\hbar k-\frac{mE^2}{2B^2}.}
$$
For a translationally invariant dispersion, the <group velocity> is $v_y=\hbar^{-1}\partial E_n/\partial k$. Therefore
$$
\boxed{v_y=-\frac EB,}
$$
in agreement with the <electric-cross-magnetic-field drift> $\mathbf E\times\mathbf B/B^2$.
In the rectangle, $k=2\pi r/L_y$ as before. The energy now changes with $r$, so the guiding-centre degeneracy is lifted, while the number of states in each tilted band remains approximately $|qB|A/(2\pi\hbar)$. Adjacent states within a formerly degenerate level have the <electric-field splitting of a Landau level in a rectangle>
$$
\boxed{\Delta E_{\rm electric}=\frac{2\pi\hbar}{L_y}\left|\frac EB\right|.}
$$
Thus this is the ground-to-first-excited gap when the lowest Landau band contains at least two allowed guiding centres and $\Delta E_{\rm electric}<\hbar\omega_c$. Without that implicit weak-field or large-sample condition, the exact full spectral gap is
$$
\boxed{\Delta E=\min\!\left\{\frac{2\pi\hbar|E|}{|B|L_y},\,\hbar\omega_c\right\}.}
$$
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