see 3 is correct but not 6,5,4,3,2,1 why reason is
first for 6 , given no of vertices is 6 in simple graph 'n' vertices maximum degree is n-1 means as given 6 vertices and one vertex should have maximum 5 edges so 6 edges not possible in case of minimum
second for 5 , only degree sequence possible is 5,3,3,3,3,3 so in this graph minimum degree is 3 whose removal makes graph disconnected
third for 4 ; degree sequence possible is 4,4,4,4,2,2 , so in this graph minimum degree is 2 whose removal makes garph disconnected
forth for 3, same above 5,3,3,3,3,3 so min degree is 3
fifth for 2 ; degree sequence possible is 4,4,4,4,2,2 same above
sixth for 1; no degree sequence possible
so total 2 graph possible 1st have degree sequence is 4,4,4,4,2,2 let named is G1
2nd have degree sequence is 5,3,3,3,3,3 let named is G2
in G1 '2' edges removal graph disconnected and G2 '3' edges removal graph disconnected
now what should be the answer see question "minimum number of edges whose deletion from graph G is always guarantee" now removing min edge 2 work only for G1 not G2 bu removing min edge 3 always guarantee for G1 as well as G2 so ans is 3 minimum edge