128 is the correct answer.
maximum 2 address instructions possible = 2^4 = 16,
but only 14 double address instructions used, so 16-14 = 2 double address instructions remains. Now for next 6 bit, we can have 2^6 = 64 combinations
=>2*64 = 128 comb for one address instructions.
So,answer is 128 only.