physical memory 32 bit
cache memory 256 KB = 2^18
4 way set associative
cache block size 16B = 2^4
so block offset bit =4
no of blocks = 2^14
no of sets = 2^14 / 4 = 2^12
set offset bits = 12 bits
so tag bits = 32 - 12 - 4 =16 bits.
size of comparator=no of tag bits
size of comparator needed is 16 bits only.
(all options are wrong,basically they wanted to know the tag memory size)
no of tag bits in set associative cache is = 16
size of tag bits is = 16* 2^14 = 2^18.