Find the greatest value of b so that 30a68b (a > b) is divisible by 11.

Given, 30a68b is divisible by 11.

$$\Rightarrow$$  (3 + a + 8) - (0 + 6 + b) will be multiple of 11.

$$\Rightarrow$$  (a - b + 5) will be multiple of 11.

$$\Rightarrow$$  (a - b + 5) = 0 or (a - b + 5) = 11 or (a - b + 5) = 22 so on.

$$\Rightarrow$$  a - b = -5  or  a - b = 6  or  a - b = 17  so on.

Since a > b, a - b cannot be negative.

Both a and b are digits, so a - b cannot be a two digit number.

The only possibility is 

a - b = 6

The pairs satisfying the above equation are (9,3), (8,2), (7,1), (6,0).

The greatest value of b can be 3.

Hence, the correct answer is Option C