Approach should be finding number of bits in the set offset.
Set bits = log(CS/BS/k) = log(64K/256/4) = 6
Now see the addresses, the address that matters to us is C8, 68,28,48
First 6 bits from the LSB side are
001000, 101000, 101000, 001000
clearly A2 and A3 are mapped to same cache sets.