Didn't use any formula, instead went by the naive method.
Superkeys for A : A, AB, AC, AD, ABC, ABD, ACD, ABCD. So n(A) = 8
Superkeys for BC : BC, BCD, ABC, ABCD. So n(BC) = 4
Here, n(A $\cap$ BC) = 2 as ABC and ABCD are common superkeys
So taking union, we get
n(A $\cup$ BC) = 8 + 4 - 2 = 10