Answer:
10 bits
Explanation:
Cache size = 32 KB
We were also given that it is 2 way set associative
Block size = 16 bytes = 2^4 = 4 bits for offset
Number of blocks = cache size / block size = (32*2^10) / 16
=2^11
Hence, index = number of blocks / 2 (since 2 way associative)
=(2^11)/2 = 2^10 = 10 bits for index
