 Solution

For hyperfine coupling for a spin-one nucleus, let the electron spin be $s=1/2$ and the nuclear spin $I=1$. First couple $\mathbf{j}=\mathbf{L}+\mathbf{s}$, giving $j=L\pm 1/2$, then $\mathbf{F}=\mathbf{j}+\mathbf{I}$, giving $F=|j-I|,|j-I|+1,\dots ,j+I$.

For each $F$, the projection is $m_F=-F,-F+1,\dots ,F$; electron projections similarly satisfy $m_j=-j,\dots ,j$. Summing $2F+1$ gives 18 states for $L=1$ and 30 for $L=2$, agreeing with $(2L+1)(2s+1)(2I+1)$.

Notation qualification: if the printed leading “2” denotes the principal quantum number, a hydrogenic $2d$ bound state does not exist because $L\le n-1$; a physical $d$ state needs $n\ge 3$. If a doublet term symbol ${}^{2}P$ or ${}^{2}D$ was intended, its leading 2 instead means $2s+1$, and the angular-momentum table applies without fixing $n$. The answers above use the explicitly specified $L$ values, while retaining this ambiguity.^[ref]