= Solution
Write $b=(\ell+1)a$, $R(r)=r^\ell e^{-r/b}$ and $Y(\theta,\phi)=\sin^\ell\theta e^{i\ell\phi}$, for nonnegative integer $\ell$. The <Laplace-Beltrami operator>[angular Laplacian] satisfies $\Delta_{S^2}Y=-\ell(\ell+1)Y$. Direct radial differentiation gives
$$
R''+\frac2rR'-\frac{\ell(\ell+1)}{r^2}R=\left(\frac1{b^2}-\frac{2(\ell+1)}{br}\right)R.
$$
Because $(\ell+1)/b=1/a$, the $1/r$ contribution cancels the <Coulomb potential> in the <Hamiltonian>. Therefore
$$
\boxed{H(RY)=-\frac{\hbar^2}{2mb^2}RY,\qquad E=-\frac{\hbar^2}{2m(\ell+1)^2a^2}.}
$$
The <volume element> contributes $r^2\sin\theta$. The squared state factors into radial, polar and uniform azimuthal densities, so the angular <normalization> cancels in every <expectation value>[radial expectation]. With $I_j=\int_0^\infty r^je^{-2r/b}\,dr$,
$$
\boxed{\mathbb E[r^k]=\frac{I_{2\ell+2+k}}{I_{2\ell+2}}.}
$$
The supplied integration-by-parts recurrence gives $I_{j+1}=(j+1)bI_j/2$. Hence
$$
\boxed{\mu=\mathbb E[r]=(\ell+\tfrac32)b,\quad\mathbb E[r^2]=(\ell+2)(\ell+\tfrac32)b^2,\quad\operatorname{Var}(r)=\frac{2\ell+3}{4}b^2.}
$$
In particular $\operatorname{Var}(r)/\mu^2=(2\ell+3)^{-1}$. Put $h=(2\ell+3)^{-1/3}$. <Chebyshev's inequality> gives
$$
\boxed{\mathbb P(|1-r/\mu|>h)\leq\frac1{(2\ell+3)h^2}=h.}
$$
For the polar variable set $J_j=\int_0^\pi\sin^j\theta\,d\theta$. The polar density is proportional to $\sin^{2\ell+1}\theta$, so
$$
\boxed{\mathbb E[\cos^2\theta]=1-\frac{J_{2\ell+3}}{J_{2\ell+1}}=\frac1{2\ell+3}.}
$$
An exact expression for the requested tail, as well as its useful estimate, is
$$
\boxed{\mathbb P(|\cos\theta|>h)=\frac{\int_h^1(1-z^2)^\ell\,dz}{\int_0^1(1-z^2)^\ell\,dz}\leq\frac{\mathbb E[\cos^2\theta]}{h^2}=h.}
$$
The inequality is <Markov's inequality> applied to $\cos^2\theta$. If a finite expression is desired, expand $(1-z^2)^\ell$: the numerator is $\sum_{j=0}^\ell(-1)^j\binom\ell j(1-h^{2j+1})/(2j+1)$ and the denominator is the same sum with $h=0$.
By the <union bound>, with <probability> at least $1-2h$ both $|r/\mu-1|\leq h$ and $|\cos\theta|\leq h$. The height above the equatorial plane is then at most $(1+h)\mu h$. The cylindrical radius differs from $\mu$ by at most $\mu h+(1+h)\mu h^2$. Thus the distance from the circle of radius $\mu$ in the equatorial plane is $O(\mu h)$ with <probability> tending to one. This is relative concentration around a growing circle; it does not assert an absolutely shrinking thickness. The <probability concentration of a circular Coulomb eigenstate> is a precise version of the large-angular-momentum orbit picture.
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