2022 AMC 10B Problem 3

There are $9\cdot10=90$ choices for the hundreds and tens digits. Once those two digits are fixed, the units digit must have whichever parity makes the total number of even digits odd. There are always $5$ digits of the required parity (including $0$ among the even digits). Therefore, the number of integers is $90\cdot5=450.$ Thus, the answer is D .