Firstly find the candidate keys for given relation R. These are :AD, BD, ED, FD
Now after finding candidate keys, check the relation for BCNF, 3NF and 2NF
For BCNF, every determinent must be super key. But this is not followed by any FD, so relation is not in BCNF.
For 3NF, if there is X-> Y dependency then Y must be prime attribute, which is satisfied by given FDs (like CH->G, G is not prime) , so relation is not in 3NF.
For 2NF, there must not be any partial dependency, but A->BC is partial dependency, so relation is not in 2NF.
According to given question, all values are atomic so relation is in 1NF.
Correct option is A.