2020 AMC 10A Problem 6

The last digit must be compatible with both evenness and divisibility by $5$

Choose the thousands, hundreds, and tens digits independently after fixing the last digit

The last digit must be $0$, because the number is divisible by $5$ and all digits are even. The thousands digit can be $2,4,6,$ or $8$, and each of the hundreds and tens digits has $5$ choices. Thus there are $4\cdot5\cdot5=100$ such integers. Thus, B is the correct answer.