Question

How many total bits are required for a direct- mapped cache with 256 KB of data and 8- word block size, assuming a 32-bit address?

Answer 1

it's direct mapped (associativity=1), it holds 16k of data, and our blocksize is 64 bytes. so we have 16k/64 = (16*1024)/64 = 256 sets. log2(256) = 8, so we need 8 bits for the index.