I think answer is A .
Question is that how much memory is needed to store each user's access data so its only the memory reqd for read/write information which is asked. For 2^16 files, there are 4 different combinations of read(R) write(W). Read, write, no read, no write.
So bits reqd for 1 file to store access info=2
Bits reqd for 2^16 files=(2^16)×2 bits=128Kb
Converting this bits into bytes we get 128Kb/(8bits/Byte)
=16KB