For a 2 address instruction Instruction bit is divided as
4 bit-Opcode
6 bit- address1
6 bit- address2
Total instruction possible with 2 address =2^4=16
For one address instruction(given opcode expanding)
No of one address instruction= No of free 2 address opcode×2^6(Why? Only one address is used in one address so one of the 2, 6 bit address which was used for 2 address instruction can now be used to calculate new instructions in one address)
192(given)=No of free opcode×2^6
No of free opcode=192/64=3
Total opcode=16
No of two address instruction=16-3=13