look, main memory 2^28
so 28 bit
now cache size 2^13
cache line size 2^4
no of block = 2^9
2 way set associative
so no of set 2^8.
now in 28 bit,
tag(16) | set(8) | word offset (4) |
look at the option 7 digit hex given
so total 28 bits.
last 4 bit for word offset.
for determinning set we have to look on the two digits before the last digit.
for set 0
as we can see set has 8 bits so put 00000000 which 00 in hex.
and only (CFED00B)16 this matched .
option C.