The problem is Solved by Using Strongly Connected component by Calling DFS.
Call DFS for each option according to there start and Finish time.(<U,V> meaning U is starting time and V is finish time of vertex)
1) <(1,6) (2,5) (3,4) (8,10)>: This option having 2 connected component as 1/6,2/5,3/4 and 8/10 so not connected graph
2) <(6,7) (2,5) (3,4) (8,9)>: In this graph having 3 connected component like (2/5,3/4) as one component,(6/7) is second and (8/9) is third component so it also not connected graph.
3) <(4,5) (2,8) (1,7) (3,6)>: In this graph is connected as having single connected component as 1/8,2/7,3/6,4/5.
4) <(7,8) (1,2) (5,6) (3,4)> : graph having 4 connected component as (1/2),(3/4),(5/6),(7/8).
option C is correct.