Answer
For each value of $n$, there are $~~n^2~~$ possible values of $m_l$
Work Step by Step
For each value of $n$, the values of $l$ can be $~~l = 0, 1, 2,...,(n-1)$
$n~~$ values of the orbital quantum number $~l~$ are possible.
For each value of $l$, the values of $m_l$ can be $~~m_l = 0, \pm 1, \pm 2,...,\pm l$
$2l+1~~$ values of the orbital magnetic quantum number $~m_l~$ are possible.
For each $n$, we can find the number $N$ of possible values of $m_l$:
$N = 1+3+5+...+[2(n-1)+1]$
$N = \sum_{0}^{n-1}(2i+1)$
$N = \sum_{0}^{n-1}2i+n$
$N = 2~\sum_{0}^{n-1}i+n$
$N = (2)~[\frac{(n-1)(n)}{2}]+n$
$N = (n-1)(n)+n$
$N = (n^2-n)+n$
$N = n^2$
For each value of $n$, there are $~~n^2~~$ possible values of $m_l$
