Out of 11 bits, 3 bits are for opcode in 2-Address. So 2^3=8 instructions(2-Address) are possible.
But 5 are being used. So 3 are free.
We use these 3 for 1-address instruction by using 4 bits of 1 of the addresses i.e. 3*16=48
Now out of 48 1-address instructions, 32 are being used. So we are left with 16.
Again use these 16 for 0-address instruction by using 4 bits of address i.e. 16*16=256