A vertex will have n-1 edges going out of it. For node to be of degree two, 2 of these n-1 edges need to be selected.
$P=\binom{n-1}{2}\left ( p \right )^{2}\left ( 1-p \right )^{n-3}$
We have n nodes, so expected no of nodes of degree 2
E= n*P
$E= n\binom{n-1}{2}\left ( p \right )^{2}\left ( 1-p \right )^{n-3}$