= Solution
The <linear confinement correction to Coulomb bound states> uses $H'=br$, and its <first-order energy correction> in a normalized Coulomb <eigenstate> is $b\langle r\rangle$. Replace $a_0$ by $a_q$ in the <hydrogen> <wavefunctions>. The normalized <spherical harmonics> remove the angular <integral>, leaving
$$
\langle r\rangle_{nl}=\int_0^\infty r^3|R_{nl}(r)|^2\,dr.
$$
For $1s$, the <integral> of $r^3e^{-2r/a_q}$ yields
$$
\langle r\rangle_{1s}=\frac4{a_q^3}\frac{3!}{(2/a_q)^4}=\frac32a_q.
$$
For $2s$, put $u=r/a_q$ and square the full radial <polynomial>, including its cross term:
$$
\langle r\rangle_{2s}=\frac{a_q}{2}\int_0^\infty u^3\left(1-u+\frac{u^2}4\right)e^{-u}du
=\frac{a_q}{2}\left(3!-4!+\frac{5!}4\right)=6a_q.
$$
For $2p$, the same calculation gives
$$
\langle r\rangle_{2p}=\frac{a_q}{24}\int_0^\infty u^5e^{-u}du=5a_q.
$$
Consequently the first two Coulomb levels become, to first order,
$$
\boxed{\begin{aligned}
E_{1s}&=-\frac{m_qa^2}{4\hbar^2}+\frac32ba_q+O(b^2),\\
E_{2s}&=-\frac{m_qa^2}{16\hbar^2}+6ba_q+O(b^2),\\
E_{2p,m}&=-\frac{m_qa^2}{16\hbar^2}+5ba_q+O(b^2),\qquad m=-1,0,1.
\end{aligned}}
$$
Thus the second level splits into a singlet orbital level and a triply degenerate orbital level, with $E_{2s}-E_{2p}=ba_q+O(b^2)$. For the confining sign $b>0$, $2p$ lies lower than $2s$: its smaller mean radius gives the smaller positive correction. The expansion assumes these corrections and the induced couplings to other principal shells are small compared with the relevant Coulomb <energy> gaps; $ba_q^2/a\ll1$ is the governing small parameter for fixed low $n$.
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