Let us consider an array of 11 element which are sorted.
10 20 30 40 50 60 70 80 90 100 110
Now apply binary search in the given array .For convenience we are using Binary search tree.
For first element (which is root) array will be
(10 20 30 40 50 ) 60 (70 80 90 100 110)
Now binary search continue either in lower or in upper halve for searching .
For lower halve (10 20) 30 (40 50).
For upper halve (70 80) 90 (100 110) and continued in same way.

So to search middle element (i.e 60 ) it will take 1 comparison .For 30 or 90 it will take 2 comparison .For 20 ,3 comparison and so on.
So avg No of comparison=$\large \frac{1+(2+2)+(3+3+3+3)+(4+4+4+4)}{11}=3$
So option A is correct.