Given,
cache size = 256KB
PA = 36 bits
set associativity = 16
So, assume block size is $2^{x}$
Then, no of cache blocks = $\frac{2^{18}}{2^{x}}$
no of sets in cache would be = $2^{18-x-4}$
So now the PA split up would be
t + 14-x + x =36 (Assume t bits for tag)
t = 22 bits.
Check this question for reference:
https://gateoverflow.in/39622/gate2016-2-32?show=39622#q39622