In question, it’s told that the size of the physical address space = 32 GB, what I thought is the size of physical memory is 32GB. Because of which I started to find the number of blocks in physical memory,
Number of blocks = Physical address space/ block size.
As it’s not mentioned explicitly that cache block size and main memory size are different, so we can assume that both are equivalent. So, what I got is the number of blocks in physical memory is 2^27 which can be accessed using 27 bits. Hence, my total instruction length came out to be 27 bits. I found set bits and word bits correctly, but the remaining was only 15 bits which were finally assigned to the Tag section of instruction.
Please correct me where I am getting wrong.