An ISP having an address block 202.16.0.0/15 assigns a block of 6000 IP addresses to a client, using the classless internet domain routing (CIDR) super-netting approach. Which of the following address blocks can be assigned by the ISP?

1. Determine required block size: The client requires $6000$ IP addresses. In CIDR, the number of addresses allocated in a block must be a power of $2$.

   $$\lceil \log_2(6000) \rceil = 13 \implies 2^{13} = 8192$$

   The prefix length is therefore $32 - 13 = 19$, meaning the assigned block must have a /19 mask.

2. ISP's Address Range: The ISP owns 202.16.0.0/15:

   • Start address: 202.16.0.0

   • End address: 202.17.255.255

3. Valid /19 Block Boundaries: In a /19 block, the first two octets are fixed per block part, and the 3rd octet has $19 - 16 = 3$ network bits. The step size (block increment) in the 3rd octet is $2^{8 - 3} = 2^5 = 32$. Thus, the valid start values for the 3rd octet are multiples of $32$: $0, 32, 64, 96, 128, 160, 192, 224$.

4. Evaluate Options:

   • (A) 202.16.0.0/19: 3rd octet is $0$ (multiple of $32$), within ISP range (202.16.0.0 to 202.16.31.255). Valid.

   • (B) 202.17.64.0/19: 3rd octet is $64$ (multiple of $32$), within ISP range (202.17.64.0 to 202.17.95.255). Valid.

   • (C) 202.16.32.0/19: 3rd octet is $32$ (multiple of $32$), within ISP range (202.16.32.0 to 202.16.63.255). Valid.

   • (D) 202.17.24.0/19: 3rd octet is $24$, which is not a multiple of $32$. Thus, this cannot be a valid CIDR network boundary for /19. Invalid.

Hence, options A, B, and C are correct.