Question

A computer uses a memory unit with 256 k words of 32 bits each. A binary instruction code is stored in one word of memory. The instruction has four parts: an indirect bit, an operation code and a register code part to specify one of 64 registers and an address part. How many bits are there in the operation code, the register code part and the address part?

Answer 1

bits of register code part is log2(64)=6bits

bits of memory address part is log2(256k)=18bits(if it uses whole 18 bits to directly access memory).but there are also other ways to access memory,for example you appoint a base address that exist in a register and add it to an immediate number that is in the instruction word,to form an address to access memory(shorter of address bits in this way)

i cant say how many bits there are in the operation code,it  depends on the designer.