6 bit addresses ....2 address instructions are also there
so 12 bit for address and 4 bit for opcode
now let 2 address instructions be n so no.of unused instructions will be 2^4-n
now for expanding opcode we will have
(2^4-n)*2^6 on adress which will be equal to 192
(2^4-n)*2^6=192 solving it we get n as 13