Solution:

The answer is (C). Let $n$ be the number of exams passed by the student with the average of 23. Let $S$ be the sum of the grades obtained by the student before the last exam and $x$ the grade obtained in the last exam. We have:
$$\left\{\begin{array}{l} S=23 n \\ S+x=22.25(n+1) \end{array}\right.$$
Subtracting the first equation from the second we get $x=22.25-0.75 n$, that is $4 x=89-3 n$. It follows that $89-3 n$ is divisible by 4. Since $x$ is an integer greater than or equal to 18 and $n>0$, we also have $72 \leq 89-3 n<89$, from which $n=3$ and $x=20$.