Number of sets in TLB $= 256/8 = 32.$
Number of PTEs $ =2^{64} B/8KB = 2^{51} $
So, number of PTEs which can map to a TLB set $ = 2^{51}/32= 2^{46}.$
So, minimum size of TLB tag $ = 46$ bits.
We have $256$ entries in TLB each of $46$ bits. So, total tag memory size for TLB $ = 46 \times 256\; \text{bits} = 46 \times 32\; \text{bytes} = 1472\:\text{bytes.}$