Answer
32 states.
Work Step by Step
For n = 4, the value of $\mathcal{l}$ ranges from 0 to 3.
For each $\mathcal{l}$, the value of $m_{\mathcal{l}}$ goes from $-\mathcal{l}$ to $+\mathcal{l}$, for a total of $2\mathcal{l}+1$ values.
For each $m_{\mathcal{l}}$ there are two values of spin, $m_{s}$.
Thus the number of states for each $\mathcal{l}$ is $2(2\mathcal{l}+1)$.
For n = 4, the number of states is $$2(0+1)+2(2+1)+2(4+1)+2(6+1)=32 \;states$$
See the figure for the complete list of states.
We start with $\mathcal{l}=0$, and list the quantum numbers in the order $(n, \mathcal{l}, m_{\mathcal{l}}, m_s)$.
