= Solution
The <probability> of finding the electron in a spherical shell is $P(r)\,dr=4\pi r^2|\psi(r)|^2dr$, so
$$
P(r)=4\beta^3r^2e^{-2\beta r}.
$$
This <radial probability density>, rather than the density per unit volume, determines the most probable radius. Its <expectation value> is
$$
\langle r\rangle=4\beta^3\int_0^\infty r^3e^{-2\beta r}\,dr=\boxed{\frac{3}{2\beta}=\frac{3a_0}{2Z}.}
$$
For its interior maximum, differentiate $\log P=\text{constant}+2\log r-2\beta r$. The stationary condition $2/r-2\beta=0$ gives
$$
\boxed{r_{\mathrm{mode}}=\frac1\beta=\frac{a_0}{Z}.}
$$
The density per unit volume $|\psi|^2$ is instead largest at the nucleus; including the spherical-shell volume factor is essential to the requested radial <mode>.
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