The number of edges is $m$. The intersection of unions is equal to the union of the intersections. Therefore, the set of edges described contains $E_1$, since there is an $E_1$ in both terms of the intersection. Any other term intersected with $E_1$ will be a subset of $E_1$ and so is already taken care of. Therefore, we need only consider the intersections of the remaining terms. By assumption, $E_2 \cap E_3$ is empty, as is the intersection of the last terms since $V_2 \cap V_3$ is empty. This means the only intersections left to worry about are the intersection between $E_2$ and all possible edges between $V_1$ and $V_3$ (and the same with 2 and 3 switched). However the edges in $E_2$ have endpoints in $V_2$, which is disjoint from $V_3$, and so the intersection is empty. Therefore, the only edges in the intersection are in $E_1$, which has size $m$.