16,383.
32 bit architecture means word size of CPU is 32 bit, which means instruction size = 32 bit.
It is given that register memory size is 64 register, to access 1 register 6 bits are required(log 64 base2)
The size of ISA(instruction set architecture) = 45 which require 6 bit for opcode(as 5 won’t be sufficient)
So, now the feilds of instruction will have bifurcation like this,
opcode+register1+register2+Immediate value = 32 bit
Immediate value = 32 – (6+6+6) = 14
Now, Immediate value is given to be unsigned and the range of unsigned is 0 to 2^(n) – 1
So maximum unsigned value that can be represented in this instruction = 2^(14) – 1 = 16,383.