Instruction size if of 32 bits.
according to given information here we have 2 different types oprands and one opcode.
so first find register mode oprand bits, how -> given total 64 rigesters, take log(64)=6
now find opcode field bits, how -> given 45 types of instructions, take ceil{log(45)}=6
now find immediate operand field bits, how -> total bits- other field bits
= 32 - (6+6+6) (2 times 6 bcz two register oprands given)
= 14
*note: base of log is 2.