To find the number of 3-digit numbers that can be formed using the digits 0, 2, 5, 3, and 7, we need to consider the following conditions:

1. The number should be divisible by 5.
2. None of the digits should be repeated.

Let's break down the solution into two parts:

Part 1: Divisibility by 5
To be divisible by 5, the last digit of the number must be either 0 or 5. Therefore, we can only choose either 0 or 5 for the units place.

Part 2: Non-repetition of digits
Since none of the digits should be repeated, we have the following options for the remaining two places:

For the hundreds place: We can choose any of the remaining three digits (2, 3, or 7).
For the tens place: We can choose any of the remaining two digits.

Now, let's calculate the total number of possible numbers:

- For the units place: 2 options (0 or 5)
- For the hundreds place: 3 options (2, 3, or 7)
- For the tens place: 2 options (the remaining two digits)

Total number of possible numbers = 2 * 3 * 2 = 12

However, we need to consider that the hundreds place can also be 0. In this case, we have the following options:

- For the units place: 2 options (0 or 5)
- For the tens place: 2 options (the remaining two digits)

Total number of possible numbers with 0 in the hundreds place = 2 * 2 = 4

Adding the two cases together, we get a total of 12 + 4 = 16 possible numbers.

However, we need to exclude the case where the number is 000 (which is not a 3-digit number). Therefore, the final answer is 16 - 1 = 15.

Hence, the correct answer is none of the given options.