A $5$ -digit number $247X8$ is divisible by $44$. Which digit can replace $X$?

For a number to be divisible by $44$, it must be completely divisible by both $4$ and $11$ ($44=4\times 11$).
1. Divisibility rule of 4: The last two digits must be divisible by $4$. Here, the last two digits are $X8$. The choices for an even digit $X$ could be $0,2,4,6,8$ to make numbers like $08,28,48,68,88$ divisible by $4$.
2. Divisibility rule of 11: The difference between the sum of digits at odd positions and the sum of digits at even positions must be either $0$ or a multiple of $11$.
* Sum of digits at odd positions (1st, 3rd, 5th) $=2+7+8=17$
* Sum of digits at even positions (2nd, 4th) $=4+X$

Difference $=17-(4+X)=13-X$
For this difference to be a multiple of $11$, we set:

$$13-X=11⟹X=2$$

Since $X=2$ satisfies both the divisibility conditions (the last two digits $28$ are divisible by $4$), the correct option is $2$.