Answer. C. $\Theta (n^{2})$
Explanation
Question marks two key points here.
1. Sorted order is to be maintained.
2. 'n' elements to be inserted in a linked list.
Hence, for each element, we need to search it's correct position for every insertion. After the search is complete, insertion will take constatn time (i.e. mapping the correct pointers)
Now, considering the worst case where the inserting is made in last node (we'll have to scan entire list everytime)
For inserting first element, no. of elements scanned= 0
For inserting second element, no. of elements scanned = 1
For inserting third element, no. of elements scanned = 2
.
.
.For inserting nth element, no. of elements scanned = n-1
Total scans = 1+2+3+...(n-1) //Thats an AP !
Total scans = $\frac{n}{2}((n-1)+1)$ = ~$n^{2}$
We can conclude is worst case as O($n^{2}$)