= Solution
Take the <hydrogen> ground-state binding-energy magnitude as $E_B\simeq13.6\,\mathrm{eV}=2.18\times10^{-18}\,\mathrm J$, and use $a_0\simeq5.29\times10^{-11}\,\mathrm m$ as its characteristic radius. A definite volume convention is
$$
V_H=\frac{4\pi a_0^3}{3}\simeq6.21\times10^{-31}\,\mathrm{m^3}.
$$
It gives
$$
\boxed{u_H=\frac{E_B}{V_H}\simeq3.51\times10^{12}\,\mathrm{J\,m^{-3}},
\qquad\frac{u_{\rm vac}}{u_H}\simeq1.78\times10^3.}
$$
A quantum atom has no sharp outer surface. If instead its characteristic volume is taken as $a_0^3$, the estimate is $1.47\times10^{13}\,\mathrm{J\,m^{-3}}$ and the ratio is about $4.2\times10^2$. Both conventions place the two densities roughly three orders of magnitude apart at order-of-magnitude accuracy. The comparison uses the magnitude of the <binding energy>, not its negative sign relative to <ionization>.
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