Relation is in 2NF or not?
Second normal form means that each non prime attribute must not depend on a proper subset of a candidate key.
In your example, assuming that the given FDs are a cover of the FDs of the relation, you have two candidate keys, ABC and ABD. Moreover, the attribute E depends on BC, which is a proper subset of the candidate key ABC (you can verify this by computing the closure of BC, BC+, and see that E belongs to that closure).
So the relation is not in 2NF.