Given information :-
Size of cache =$32$ KB
Cache is 4 way set associative implies each set contain 4 lines .
Size of main memory =$2^{32}$ Byte
Memory word = $8$ byte
Block size =64 byte
So number of sets= $\frac{2^{15}}{64*4}=128=2^{7}$ , So 7 bits required for set number .
Tag bits $=(32-7-6)=19$
Tag bits Set bits offset
Now , each line contain $64$ byte of data and each memory word size is 8 byte . It is given each line maintain $1$ dirty bit per word .
So number of word in a line = $\frac{64}{8}=8$ .
So each line maintain $8$ dirty bits .
So total overhead per line =( data +tag bits + Dirty bit + valid bit + LRU bit)
$=(64*8 + 19+8+1+2)$
$=542$ bit
Now as the given cache is $4$ way set associative so a set consists of $4$ line .
So total size of a single set= $4*512= 2168$ bits .