IP: 200.1.2.3 - Class C IP
255.255.255.192: SM = (N/w ID + Subnet ID = 1's && Host ID = 0's)
11111111.11111111.11111111.11000000 - two subnet ID
So 22 = 4 subnets.
Each subnet will eat up 2 IPs ( 1 for N/w ID and 1 for Direct Broadcasting ).
So Answer is 4×2= 8