NO. of bits required are-
block size= 64 = $2^{6}$ (6 bits)
cache size = 4k = $2^{12}$
number of lines= $\frac{cache size}{block size}$ = $\frac{4K}{64}$ = 64 = $2^{6}$
number of set = $\frac{number of lines}{block per set}$ = $\frac{64}{4}$ = 16 = $2^{4}$ (4 bits)
Tag SET NO BLOCK OFFSET
Ans: D. 4,6