= Solution
Take the spherical harmonic $Y_{00}$ to have unit angular norm. The radial normalization condition is then
$$
1=\int_0^\infty |R(r)|^2r^2\,dr
=C^2\int_0^\infty r^2e^{-2\gamma r}\,dr
=\frac{C^2}{4\gamma^3}.
$$
Therefore
$$
\boxed{C=2\gamma^{3/2}}.
$$
Equivalently, if $R$ denotes the entire spherically symmetric wavefunction rather than the radial factor multiplying normalized $Y_{00}$, then $C=(\gamma^3/\pi)^{1/2}$.
The expected radius is
$$
\begin{aligned}
\langle r\rangle_R
&=C^2\int_0^\infty r^3e^{-2\gamma r}\,dr\\
&=4\gamma^3\frac{3!}{(2\gamma)^4}
=\boxed{\frac{3}{2\gamma}}.
\end{aligned}
$$
For the ground state, $\gamma=mq^2/\hbar^2$. The <Bohr radius> is $a_0=\hbar^2/(mq^2)=1/\gamma$, so
$$
\boxed{\langle r\rangle_R=\frac32a_0}.
$$
The mean radius is therefore of the Bohr-radius scale and is one and a half times $a_0$. This is the <radial normalization of the hydrogen ground state>.
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