Solution:

The answer is 1274. Notice that the number $a1211b$ must be divisible by 88, so it must be divisible by both 11 and by 8. We know that a number is divisible by 8 if and only if its last 3 digits are, so $11b$ must be divisible by 8. Notice that $8 \cdot 14 = 112$ is the only number of the form $11b$ divisible by 8, so $b = 2$ and $a12112$ is divisible by 8 for every choice of $a$. The divisibility criterion for 11 tells us that the difference between the sum of the digits in even position and that of the digits in odd position must be a multiple of 11, that is, that $(2+1+1)-(1+2+a) = 1-a$ must be divisible by 11. Since $a$ is a digit, hence between 0 and 9, this condition is satisfied only for $a = 1$. The solution is therefore $\frac{112112}{88} = 1274$.