Answer:
d) 152.77.0.0/19 (255.255.224.0)
Step-by-step explanation:
If it is required to use a single mask throughout a Class B network, this means that by default, we have 16 bits as network ID, and the remaining 16 bits as host ID.
If we need to set up 5 subnets and we need that the largest subnet can support a minimum of 5000 hosts, we need to divide these 16 bits from the Host ID part, as follows:
Number of hosts = 5000
We need to find the minimum power of 2 that meet this requirement:
2ˣ - 2 = 5,000 ⇒ x = 13
We have left only 3 bits from the original 16, and repeating the process, we find that we need to use the 3 bits in order to accommodate the 5 subnets,
due to with 2 bits we could only support 4.
So, we need that the address mask be, using the CIDR notation, /19, as we need that the first three bits from the third byte to be part of the network ID.
In decimal notation, it would be as follows:
255.255.224.0