Degeneracy of H atom in 3rd excited State is
1) 16
2) 10
3) 12
4) 9

Single electron species don't follow the $[n+l]$ rule. [$(n+l)$ rule is the rule to find the energy order of the subshells], thus hydrogen being a single electron specie has all its subshells of equal energy in n=4 (in the third excited state electron will be in $4^{th}$ shell). So the degeneracy of the energy levels of the hydrogen atom is $n^2$

So Degeneracy in third excited state $=4^2=16$

thus degeneracy( number of orbitals of equal energy for the electron to be excited in ) is 16 {1 s orbital+3p orbitals +5d orbitals+7orbitals}

So the correct answer is option (1).