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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

$E= n\binom{n-1}{2}\left ( p \right )^{2}\left ( 1-p \right )^{n-3}$

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