Sorry, but your answer was incorrect...
Luckily, I used a more efficient version of your method to get an answer of 118...
If the units digit is a 0, then the only two cases are that the 0 is the repeated digit, in which case the tens digit must be 0 and the hundreds digit can be 1 through 9, or the first two digits are the repeated ones, of which there are also 9 choices. Now, if the units digit is any even digit of the set \(\{ 2, 4, 6, 8 \}\), then the situation is different. If the repeated digit is to be the units digit, it can be repeated in the hundreds digit, in which case there are 9 choices of tens digit, or it can be repeated in the tens digit, in which case there are 8 choices of hundreds digit. Or, if the first two digits are the repeated ones, they can be any of 1 through 9 excluding the even units digit, for a total of 8. So, for each of 2, 4, 6 and 8, there are \(9 + 8 + 8 = 25\). So, the total number is \(2 \cdot 9 + 4 \cdot 25 = \boxed{118}\).
Thanks for your effort though!