i am assuming it is word addressable, each word is 32 bits
Computer has 256 K words ===> 18 bits required to find a word in a Memory.
Given that there are 64 registers ==> 6 bits required to identify a register uniquely
Given that instruction code stored in 1 word of memory ===> there are 32 bits in a instruction.
Instruction has four parts, 1 bit for indirect, an operation code, register code part, and address part.
==> no.of bits reserved for operation code = 32-1-6-18 = 7