The total number of three-digit numbers, with one digit repeated exactly two times, is ______.

We need to count three-digit numbers where exactly one digit is repeated exactly two times (i.e., the number has exactly two identical digits and one different digit).

Case 1: Repeated digit $$\neq 0$$ and different digit $$\neq 0$$

Choose the repeated digit: 9 choices (1-9)

Choose the different digit: 8 choices (1-9, excluding the repeated digit)

Choose which position gets the different digit: 3 positions

Count: $$9 \times 8 \times 3 = 216$$

Case 2: Repeated digit $$\neq 0$$ and different digit $$= 0$$

Choose the repeated digit: 9 choices (1-9)

The different digit is 0, which cannot be in the hundreds place

Position for 0: 2 choices (tens or units place)

Count: $$9 \times 1 \times 2 = 18$$

Case 3: Repeated digit $$= 0$$ and different digit $$\neq 0$$

Two positions have digit 0 and one has a non-zero digit

The non-zero digit must be in the hundreds place (since a three-digit number cannot start with 0)

Choose the non-zero digit: 9 choices (1-9)

The non-zero digit must go in the hundreds position: 1 way

Count: $$9 \times 1 = 9$$

Total:

$$216 + 18 + 9 = 243$$

The answer is $$\boxed{243}$$.