= Solution
The <canonical momenta> are
$$
\boxed{p_\rho=m\dot\rho,\qquad
p_\theta=m\rho^2\dot\theta+\frac{qB}{2}\rho^2.}
$$
Taking the <Legendre transform> gives
$$
\boxed{H=\frac{p_\rho^2}{2m}
+\frac1{2m}\left(\frac{p_\theta}{\rho}-\frac{qB\rho}{2}\right)^2
+\frac{k\rho^2}{2}-qE\rho\cos\theta.}
$$
Its four <Hamilton equations>, in a polar chart with $\rho>0$, are
$$
\dot\rho=\frac{p_\rho}{m},\quad
\dot\theta=\frac{p_\theta}{m\rho^2}-\frac{qB}{2m},
$$
$$
\dot p_\rho=\frac{p_\theta^2}{m\rho^3}
-\frac{q^2B^2\rho}{4m}-k\rho+qE\cos\theta,
\qquad\dot p_\theta=-qE\rho\sin\theta.
$$
Canonical <angular momentum> includes the vector-potential contribution and is not simply mechanical <angular momentum>.
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