1. We need to count three-digit numbers in which exactly one digit is repeated two times.

   So the digit pattern must be of the form: $aab,\ aba,\ baa$ where the repeated digit is one digit and the third digit is different from it.

2. Let the repeated digit be $a$ and the distinct digit be $b$, with $a \neq b.$

3. First choose the repeated digit $a$.

   Since the number is three-digit, the repeated digit can be any digit from $0$ to $9$: $10 \text{ choices for } a.$

4. Next choose the distinct digit $b$.

   It must be different from $a$, so: $9 \text{ choices for } b.$

5. For the multiset $\{a,a,b\}$, the number of distinct arrangements is: $\frac{3!}{2!}=3.$

   So at first sight, total count would be: $10 \times 9 \times 3 = 270.$

6. But this includes some invalid cases where the first digit is $0$, because a three-digit number cannot begin with $0$.

7. Count the invalid numbers:

   These happen only when $a=0$, since then the arrangements are $00b, 0b0, b00$.

   Among these, the ones starting with $0$ are: $00b,\ 0b0$ for each nonzero $b$.

   Here $b$ can be any digit from $1$ to $9$, so: $9 \times 2 = 18 \text{ invalid numbers.}$

8. Therefore the required count is: $270 - 18 = 252.$

9. Alternate check:

   • If the repeated digit is nonzero: $9$ choices for repeated digit. Then distinct digit has $9$ choices, and all $3$ arrangements are valid: $9 \times 9 \times 3 = 243.$

   • If the repeated digit is $0$: distinct digit must be $1$ to $9$ ($9$ choices), and only $b00$ is valid: $9 \times 1 = 9.$

   Total: $243 + 9 = 252.$

Hence, the total number of such three-digit numbers is: $\boxed{252}$