Well, I believe I would approach this problem like this, though somebody please correct me if I'm wrong:
For the reaction order with respect to B using trials 1 and 2:
(2)^x = 4
So x = 2, meaning it's second order with respect to B
For the reaction order with respect to A using trials 2 and 3:
(2)^y * (2)^2 = 4
So y = 0
So I would suspect that the reaction is a 0 order with respect to A. Was that what the solution set said or does it say something different? I haven't heard of the 4th line method but this seemed to work for me