Instruction size = 16 bits
Address size = 128 word =7 bits
Total Two address instruction = 16 - (7 + 7) bits = 2 bits = 4
But we use only 2 two address instruction.
Remaining 2 can be used for 1 address instruction and also 7 bits by reducing an address field.
Total 1 address instruction = 2 * 2$^{7}$ = 256
But we use 100 one address instruction.
Remaining 156 along with 7 bits by reducing an address field can be used for zero address instruction.
So Total zero address instruction = 156 * 2$^{7}$ = 19968