Final answer:
The correct subnet ID for the IPv4 CIDR address 220.93.29.167/26 is 220.93.29.128 after setting the last 6 host bits to zero, as indicated by the subnet mask /26.
Therefore, the correct answer is: option c. 220.93.29.128
Step-by-step explanation:
To determine the correct subnet ID from a given IPv4 CIDR address, you first need to understand how subnetting works.
An IPv4 CIDR address like 220.93.29.167/26 contains an IP address (220.93.29.167) and a subnet mask (indicated by /26). This subnet mask represents that the first 26 bits of the IP address are used for network identification and the remaining bits are for host identification.
The subnet ID is found by turning off all the host bits in the IP address, leaving only the network bits intact. For a subnet mask of /26, this means that the last 6 bits of the IP address are set to zero (since 32-26=6).
Converting the last octet of the IP address 167 to binary gives 10100111, and setting the last 6 bits to zero gives 10100000, which is 128 in decimal.
Therefore, the correct subnet ID for 220.93.29.167/26 is 220.93.29.128.