An organization is granted the block 130.56.0.0/16. The administrator wants to…

Final answer: 130.56.255.192 to 130.56.255.255

1. Start with the given block: 130.56.0.0/16. You need 1024 subnets.

2. Since 1024 = 2^10, you must borrow 10 bits from the host portion.

3. New prefix length = 16 + 10 = /26.

4. Addresses per subnet = 2^(32 - 26) = 64 addresses (62 usable hosts).

5. Subnets increment by 64 in the last octet. The last subnet index is 1023, so offset = 1023 × 64 = 65472, which corresponds to .255.192 in the last two octets.

6. Therefore the last subnet's network address = 130.56.255.192 and its broadcast (last address) = 130.56.255.255.