Solution:

For $\underline{74A52B1}$ to be divisible by $3$, the digit sum $19+A+B$ must be a multiple of $3$. Hence $A+B$ is $1$ less than a multiple of $3$.

For $\underline{326AB4C}$ to be divisible by $3$, the digit sum $15+A+B+C$ must be a multiple of $3$. Since $15$ is already divisible by $3$, $A+B+C$ must be divisible by $3$.

Therefore $C$ must be $1$ more than a multiple of $3$. Among the answer choices, only $1$ has that form.

Thus, A is the correct answer.