Main memory size = 32 GB = 2^35 B (Byte addressbale)
Size of physical address is = 35 bit
Let x be the set offset, y be the word offset.
No. Of set = 2^x
No. of words in block = 2^y
1 set = 8 blocks ( 8 way set associative )
2^x set contains = 2^x * 8 blocks = 2^(x+3) blocks
Cache size = No of blocks * No of words per blocks
Cache size = 2^(x+3) * 2^y = 2^(x+y+3)
Tag bits + x + y = 35 bit
x + y = 35-10 = 25 bit
Cache size = 2^(25+3) = 2^(28) B = 256 MB
Hence 256 MB is the answer