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Here given VA = 40 bit ,PA = 26 bit ,

Number of frames= 16K = 14 bits are for f and remaining bits  i.e., 12 bit are for d

Number of bits for pages p = 40 - 12 = 28 bits

No.of sets = 64 / k ==>no of bits for required = log ( 64 / k) (base 2)

Tag bits = 28 - log(64 / k)

but given tag bits = 12

28 - log(64 / k)= 12

16 = log(64 / k)

 2^16 = 64 / k

   k = 64 / 2^16 (as per question)
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