The degeneracy of hydrogen atom that has energy equal to −RH/9 is: [where RH=Rydberg constant]

Energy in H-atom E = -RH/n²
But, given that E = -RH/9
⇒ n² = 9
n = 3
As here n = 3 thus there are 3 orbits.
In general, for any energy level n, the degeneracy is given by n²
So when n = 3, the degeneracy is 9.
When n = 3, the possible combinations of (n, l, m) are (3, 2, -2), (3, 2, -1), (3, 2, 0), (3, 2, 1), (3, 2, 2), (3, 1, -1), (3, 1, 0), (3, 1, 1), and (3, 0, 0)
Hence D is the correct answer.