See total required bits in opcode for 2 address feild is 6 so we can have 2^6 instruction in which some or allof them have been in 2 address instruction which we don't know .Take it as x
now total no. of 1 address insttruction will be (2^6-x)*2^7
as given 1 address instruction is 1024 we equate the above one with this value
(2^6-n)*2^7=1024
solving this will give you 56 which is the correct answer