= Solution
For Coulomb binding, the <virial theorem> gives a typical <kinetic energy> of the order of the binding energy. Taking $p_{\rm typ}\simeq\sqrt{2m_e|E_1|}$ and using the <Heisenberg uncertainty relation> estimate $r_{\rm typ}\sim\hbar/p_{\rm typ}$ gives
$$
\boxed{r_{\rm typ}\sim\frac{\hbar}{\sqrt{2m_e(54.4\,\mathrm{eV})}}
=2.65\times10^{-11}\,\mathrm m.}
$$
The <Bohr model> gives $r_1=a_0/Z=a_0/2=2.65\times10^{-11}\,\mathrm m$ for a heavy helium nucleus, agreeing with the characteristic scale. Equivalently, minimizing the localization estimate $E(r)\sim\hbar^2/(2m_er^2)-Ze^2/(4\pi\epsilon_0r)$ gives that same scale.
This is an order-of-magnitude radius estimate, not an exact deduction of the cloud's <expected value> from the inequality alone. Numerical coefficients depend on which spread is used; $\Delta x\Delta p\ge\hbar/2$ does not assert $\langle r\rangle=\hbar/p$. For the exact hydrogenic $1s$ state, $\langle r\rangle=3a_0/(2Z)=3.97\times10^{-11}\,\mathrm m$ at $Z=2$, one and a half times the Bohr orbital radius. An electron cloud has a radial distribution, whereas the <Bohr model> assigns a definite radius.
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