The given subnet is of class B as it has $16$ Network ID bits. So, option B is ruled out.
We get the network number by the bitwise AND of IP and subnet mask. Options C and D are not answers as the second byte of IP address differs whereas second byte of subnet is all $1s.$
Let's see option A.
$158.59.195.66 \;\&\; 255.255.0.0 = 158.59.0.0$
$158.59.131.33 \;\&\; 255.255.0.0 = 158.59.0.0$
Thus we get the same network number.