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)