Not considering the electronic spin, the degeneracy of the second excited state (n = $3$) of H atom is $9$, while the degeneracy of the second excited state of ${\mathrm{H}}^{-}$ is :

Energy order of orbitals of H is decided by only principle quantum number (n) while energy order of ${\mathrm{H}}^{-}$ is decided by $(\mathrm{n}+\mathcal{l})$ rule:
Electronic configuration of $\mathrm{\prime }{\mathrm{H}}^{-}\mathrm{\prime }$ is $-1{\mathrm{s}}^2$ its energy order is decided by $\mathrm{n}+\mathcal{l}$ rule.
${\mathrm{H}}^{-}=1{\mathrm{s}}^{2}2{\mathrm{s}}^{0}2{\mathrm{p}}^0$
Its $2^{\mathrm{n}\mathrm{d}}\mathrm{ }$ excited state is 2p
And degeneracy $2\mathrm{p} \mathrm{is} 3$.