2019 AMC 10A Problem 2

Both $20!$ and $15!$ contain at least three factors of $5$ and three factors of $2$, so both are divisible by $2^3\cdot5^3=1000.$ Their difference is therefore also divisible by $1000.$ Being a multiple of $1000$ makes the last three digits $0,$ which shows that the hundreds digit is also $0.$ Thus, A is the correct answer.