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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 $=(32-7-6)=19$

 19 7 6

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 .