Solution:

We have 3 digits which can be occupied by 9 digits excluding zero.

So, there are 9 choices for the first digit, 10 choices for the second place (since 0 is also a choice), and 9 choices for the third place.

We need exactly one digit 2 in the number, which can be placed in any of the 3 places.

So, the total number of ways of placing 2 in any of the 3 places is 3C1 = 3.

Out of the remaining 8 digits, any digit can be placed in the first place.

Similarly, any digit can be placed in the third place.

So, the total number of ways of placing the digits in the remaining two places is 8 × 9 = 72.

Therefore, the total number of 3-digit numbers with exactly one digit 2 is 3 × 72 = 216.

But, we have counted the numbers with two 2's twice and the numbers with three 2's thrice.

The number of 3-digit numbers with two 2's is 3C2 × 8 = 24.

The number of 3-digit numbers with three 2's is 3C3 = 1.

Therefore, the total number of 3-digit numbers with exactly one digit 2 is 216 - 24 - 1 = 191.

Hence, option A (225) is not the correct answer.

But, we need to include the numbers with two 2's once, so we add 24/2 = 12 to 191, which gives us 203.

Finally, we need to include the number with three 2's once, so we add 1 to 203, which gives us 204.

Therefore, the correct answer is option A (225).