Total cache size $= 256\ KB$
Cache block size $=32\text{ Bytes}$
So, number of cache entries $=\dfrac{ 256\ K}{32}=8\ K$
Number of sets in cache $=\dfrac{ 8\ K}{4}=2\ K$ as cache is $4\text{-way}$ associative.
So, $\log(2048) = 11\ \text{bits}$ are needed for accessing a set. Inside a set we need to identify the cache entry.
No. of memory block possible $=\dfrac{\text{Memory size}}{\text{Cache block size}}$
$=\dfrac{2^{32}}{32} = 2^{27}$.
So, no. of memory block that can go to a single cache set
$=\dfrac{2^{27}}{2^{11}}$
$=2^{16}.$
So, we need $16\text{ tag bits}$ along with each cache entry to identify which of the possible $2^{16}$ blocks is being mapped there.
Correct Answer: $C$