Using options
F has E as direct neighbor and min. path is 3 only so eliminate d as it has 5 cost for F->E
F has C as direct neighbor,so it know to reach F,it can not be infinite so eliminate a option.
Now b and c differ only in A entry,and after two iterations E has path to reach A which it will share with F,so F knows how to reach A and with 6 length path.
So this eliminates b
c should be correct answer here.