Answer: 192.168.1.0/32

## Explanation For VPC peering connections, the CIDR blocks of the peered VPCs must **not overlap**. The development VPC has CIDR block `192.168.0.0/24`, which covers IP addresses from `192.168.0.0` to `192.168.0.255`. Let's analyze each option: **A. 10.0.1.0/32** - This is a single IP address (`10.0.1.0`). It doesn't overlap with `192.168.0.0/24`, so it's valid. However, it's not the smallest CIDR block. **B. 192.168.0.0/24** - This is **identical** to the development VPC's CIDR block, so it would overlap completely. This is **invalid** for VPC peering. **C. 192.168.1.0/32** - This is a single IP address (`192.168.1.0`). This address is outside the range of `192.168.0.0/24` (which only covers `192.168.0.0-255`). This is valid and is a `/32` CIDR block (smallest possible). **D. 10.0.1.0/24** - This covers IP addresses from `10.0.1.0` to `10.0.1.255`. It doesn't overlap with `192.168.0.0/24`, so it's valid, but it's a `/24` block (256 addresses), not the smallest. **Key Points:** 1. `/32` is the smallest possible CIDR block (single IP address) 2. `192.168.1.0/32` doesn't overlap with `192.168.0.0/24` because `192.168.1.0` is in a different subnet 3. The question asks for the **SMALLEST** valid CIDR block, which means we want the highest subnet mask value (smallest address space) Therefore, **C. 192.168.1.0/32** is the correct answer as it's the smallest valid CIDR block that doesn't overlap with the development VPC's CIDR range.

Author: LeetQuiz Editorial Team