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.
Total number of distinct cache entries $=\dfrac{2^{32}}{\text{cache entry size}}=\dfrac{2^{32}}{32}=2^{27}$
Out of this $2^{27},$ each set will be getting only $\dfrac{2^{27}}{2^{11}}=2^{16}$ possible distinct cache entries as we use the first $11 \text{ bits}$ to identify a set. So, we need $16$ bits to identify a cache entry in a set, which is the number of bits in the tag field.
Size of cache tag directory$=\text{Size of tag entry}\times \text{Number of tag entries}$
$=16 +(2+1+1) \text{ bits (2 valid, 1 modified, 1 replacement as given in question)}\times 8\ K$
$= 20\text 8 = 160\text{ Kbits}$
Not needed for this question, still:
Valid bit: Tells if the memory referenced by the cache entry is valid. Initially, when a process is loaded all entries are invalid. Only when a page is loaded, its entry becomes valid.
Modified bit: When processor writes to a cache location its modified bit is made $1.$ This information is used when a cache entry is replaced- entry $0$ means no update to main memory needed. Entry $1$ means an update is needed.
Replacement bit: This is needed for the cache replacement policy. Explained in the below link:
https://www.seas.upenn.edu/~cit595/cit595s10/handouts/LRUreplacementpolicy.pdf
Correct Answer: $A$