= Solution
The <radial probability density> includes the spherical-shell volume factor:
$$
p(r)=4\pi r^2|\psi_0(r)|^2=4\alpha_0^3r^2e^{-2\alpha_0r},\qquad r\geq0.
$$
Consequently the radial <expectation value> is
$$
\begin{aligned}
\mathbb E_{\psi_0}(r)&=4\alpha_0^3\int_0^\infty r^3e^{-2\alpha_0r}\,dr\\
&=4\alpha_0^3\frac{3!}{(2\alpha_0)^4}=\boxed{\frac3{2\alpha_0}=\frac{3\hbar^2}{2mKZ}.}
\end{aligned}
$$
Keeping $m$ and $K$ fixed, the <hydrogenic ground-state radius scaling> is therefore \b[$\langle r\rangle\propto1/Z$]: the $Z=2$ state has half the mean radius of the $Z=1$ state. Equivalently, with $a_*=\hbar^2/(mK)$, the mean radii are $3a_*/2$ and $3a_*/4$. For the physical electron-nucleus <Coulomb potential>, $a_*$ is the <Bohr radius> of the fixed-nucleus <hydrogen atom> model.
\Image[/past-exam-of-the-quantum-theory-course-of-the-university-of-oxford/2015-hydrogenic-radius.png]
{title=<Radial probability densities> for the one-electron $Z=1$ and $Z=2$ <Coulomb bound states>; dashed lines mark their mean radii}
{description=The horizontal coordinate is $s=r/a_*$. The normalized densities are $4Z^3s^2e^{-2Zs}$, so their means are $3/(2Z)$.}
The exact $Z=2$ comparison in this one-electron model describes the <hydrogen-like ion> $\mathrm{He}^+$. A neutral <helium atom> has two <electrons>; its exact size also depends on their mutual <Coulomb interaction> and is not determined by this one-electron calculation. The stated half-radius conclusion is the prediction of the model supplied in the paper.
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