Answer
See explanation. The ten lowest energy levels for electrons in neon have n = 1 or n = 2.
Work Step by Step
For n = 1, $l$ can only be 0, and $m_l$ can only be 0.
Here are the possible $(n, l, m_l,m_s)$ pairs for n =1:
$$(1,0,0,+\frac{1}{2}) (1,0,0,-\frac{1}{2})$$
For n = 2, $l$ can take on the values 1 and 0.
For $l=1$, $m_l$ can be $\pm l$.
Here are the possible $(n, l, m_l,m_s)$ pairs for n =1:
$$(2,1,0,+\frac{1}{2}) (2,1,0,-\frac{1}{2})$$
$$(2,1,1,+\frac{1}{2}) (2,1,1,-\frac{1}{2})$$
$$(2,1,-1,+\frac{1}{2}) (2,1,-1,-\frac{1}{2})$$
$$(2,0,0,+\frac{1}{2}) (2,0,0,-\frac{1}{2})$$
Those cover the 10 electrons in neon.