The correct option is D 36
For 'h' subshell, azimuthal quantum number 'l' = 5.
We know that, l, ranges from 0 to n−1 for a given principal quantum number 'n'.
Hence, the minimum possible value of of n to have ‘h' subshell is n=l+1=5+1=6
The number of orbital present =62=36