The IP address
220.127.116.11 falls in Class B. In class B, 16 bits are of Network ID and other 16 bits are of Host ID.
1 0 [This 2 bits are fixed)
| Network ID (16 bits)
|| Host ID (16 bits)
As first 2 bits are fixed we are left only with 14 bits in NID. Therefore, Number of networks through Class B = 2^14 = 2^4 * 2^10 = 16K
Similarly, about Host ID, 2^16. But as two IP's from each class are allotted for Direct Broadcasting (here,
18.104.22.168) and one for Network ID (here,
Hence, 2^16 - 2.