Here we will check options one by one, although the explanation seems to me long but once you will get how to check then yo will able to handle ;
considering option (A)
1) 11001100.11001100.11001100.10000000/11111111.11111111.11111111.11000000 Network bits-26 (as 26 bits are 1 in the subnet mask) so sub-network id = 204.204.204.128 . 6 bits ‘0’ in subnet mask,i.e. 64(i.e. 2^6) -2 = 62 hosts.
2) 11001100.11001100.11001100.00000000/11111111.11111111.11111111.10000000 Network bits-25 (as 25 bits are 1 in the subnet mask) so sub-network id = 204.204.204.0 . 7 bits ‘0’ in subnet mask,i.e. 128(i.e.2^7) -2 =126 hosts.
3) 11001100.11001100.11001100.01000000/11111111.11111111.11111111.10000000 Network bits-25 (as 25 bits are 1 in the subnet mask) so sub-network id = 204.204.204.0 . 7 bits ‘0’ in subnet mask,i.e. 128(i.e.2^7) -2 = 126 hosts .
Though the networks are divided into subnets containing 62,126,126 hosts which satisfy our minimum criteria of 50,50,100. But, two subnets have same network ID thus,they are not different.Thus,this division is only of two subnets but we need 3 ,so A option is incorrect.
consider option (B)
1) 11001100.11001100.11001100.00000000/11111111.11111111.11111111.11000000 Network bits-26 (as 26 bits are 1 in the subnet mask) so sub-network id = 204.204.204.0 . 6 bits ‘0’ in subnet mask,i.e. 64(i.e. 2^6)-2=62 .
2) 11001100.11001100.11001100.11000000/11111111.11111111.11111111.10000000 Network bits-25 (as 25 bits are 1 in the subnet mask) so sub-network id = 204.204.204.128 . 7 bits ‘0’ in subnet mask,i.e. 128(i.e.2^7)-2=126 hosts
3) 11001100.11001100.11001100.01000000/11111111.11111111.11111111.10000000 Network bits-25 (as 25 bits are 1 in the subnet mask) so sub-network id = 204.204.204.0 . 7 bits ‘0’ in subnet mask,i.e. 128(i.e.2^7)-2=126 hosts .
Though the networks are divided into subnets containing 62,126,126 hosts which satisfy our minimum criteria of 50,50,100. But, two subnets have same network ID thus,they are not different.Thus,this division is only of two subnets.hence incorrect
consider option (C)
1) 11001100.11001100.11001100.10000000/11111111.11111111.11111111.10000000 Network bits-25 (as 25 bits are 1 in the subnet mask) so sub-network id == 204.204.204.128 . 7 bits ‘0’ in subnet mask,i.e. 128(i.e. 2^7) -2 =126 hosts.
2) 11001100.11001100.11001100.11000000/11111111.11111111.11111111.11000000 Network bits-26 (as 25 bits are 1 in the subnet mask) so sub-network id == 204.204.204.192 . 6 bits ‘0’ in subnet mask,i.e. 64(i.e.2^6)-2=62 hosts.
3) 11001100.11001100.11001100.11110000/11111111.11111111.11111111.11000000 Network bits-26 (as 26 bits are 1 in the subnet mask) so sub-network id == 204.204.204.192 . 6 bits ‘0’ in subnet mask,i.e. 64(i.e.2^6) -2=62 hosts.
Though the networks are divided into subnets containing 62,62,126 hosts which satisfy our minimum criteria of 50,50,100. But, two subnets have same network ID thus,they are not different.Thus,this division is only of two subnets. hence incorrect
consider option (D)
1) 11001100.11001100.11001100.10000000/11111111.11111111.11111111.10000000 –> Network bits-25 (as 25 bits are 1 in the subnet mask) so sub-network id == 204.204.204.128 . 7 bits ‘0’ in subnet mask,i.e. 128(i.e. 2^7)-2=126 hosts.
2) 11001100.11001100.11001100.01000000/11111111.11111111.11111111.11000000 –> Network bits-26 (as 25 bits are 1 in the subnet mask) so sub-network id == 204.204.204.64. 6 bits ‘0’ in subnet mask,i.e. 64(i.e.2^6)-2=62 hosts.
3) 11001100.11001100.11001100.00000000/11111111.11111111.11111111.11000000 –> Network bits-26 (as 26 bits are 1 in the subnet mask) so sub-network id == 204.204.204.0 . 6 bits ‘0’ in subnet mask,i.e. 64(i.e.2^6)-2=62 host bits. This satisfies the minimum criteria of 50,50 and 100 hosts and all subnet IDs are different .
THUS , option (D) is correct. satisfying the requirements.