$32$ bit address is used for accessing the cache.
It is given that cache is Set-Associative.
The address bits get split as follows:
Block Size $= 64 B \implies$ Block offset $= 6\; bits.$
Given that Tag field width $= 22\; bits.$
Therefore, width of Set Index field $= 32-22-6 = 4 \implies 2^4\text{ = 16 sets in the cache.}$
Cache size is $2 KB$ and Block size $= 64 B$ $\implies 2^5 \text{ = 32 blocks present in the cache. }$
$16$ sets contain $32$ blocks $\implies 2$ blocks per set or associativity $=2.$
Correct Answer: $2$