Let $X_i$ be 1 if the $i$th edge is in $E’$ and 0 otherwise.
For an edge to be in $E’$, both of its vertices must be in $V’$
$ Pr(X_i=1) = \frac{1}{4} $
Now, we need to check how many edges in $E$ are in $E’$ based on the vertices in $V’$
$ \mathbb{E}(X) = \sum_{i=1}^{m} \mathbb{E}(X_i) $
$ \mathbb{E}(X) = \sum_{i=1}^{m} Pr(X_i=1) $
$ \mathbb{E}(X) = \sum_{i=1}^{m} \frac{1}{4} $
$ \mathbb{E}(X) = \frac{m}{4} $