since instruction size is not give, assuming 1 W as instruction size. Since word size not given assuming 1W = 8 bits
total 2 addr opcodes possible = $2^{8-2i}$ out of these x are taken so, free = $2^{8-2i}-x$
total 1 addr opcodes possible = $(2^{8-2i}-x)*2^i$, out of these y are taken so, free = $(2^{8-2i}-x)*2^i-y$
total 0 addr opcodes possible = $((2^{8-2i}-x)*2^i-y)*2^i$
Now to maximise the total 0 addr opcodes possible, we need to leave as many FREE opcodes as possible in 2 addr and 1 addr opcodes pool.
So put x = 1, y = 1 (need to take minimum one 2 addr and one 1 addr, not mentioned in question, so assuming)
maximum 0 addr possible =
$((2^{8-2i}-1)*2^i-1)*2^i$ OR
$((2^{8-2i})*2^i)*2^i$ (if question allows to have zero 2 addr and zero 1 addr)
