3. (10 points) $\overline{\mathrm{abc}}$ is a three-digit number. If $a$ is an odd number, and $\overline{\mathrm{abc}}$ is a multiple of 3, then the smallest $\overline{\mathrm{abc}}$ is
【Answer】Solution: To make $\overline{\mathrm{abc}}$ the smallest, the smallest hundred's digit is 1, then the ten's digit is 0, so this number is $\overline{10 \mathrm{c}}$. Since $\overline{\mathrm{abc}}$ is a multiple of 3, we can conclude: $1+0+c$ is a multiple of 3,
Therefore, the smallest $c$ is 2,
Thus, $\overline{\mathrm{abc}}$ is 102.
So the answer is: 102.
【Comment】This problem examines the flexible application of the characteristics of numbers divisible by 3, the key is to determine the digits in the hundred's and ten's places.