if G has 1 or 2 vertices , the result is true . if G has at least three vertices ,
now given is e<=3v-6 <->2e<=6v-12. now if the degree of every vertex were atleast six , then by handshaking theram 2e=summation of degree of all vertices. so from here we can conclude that 2e>=6v , but given 2e<=6v-12 it means there is contradiction , so we must have at least 1 vertex of of degree not greater than 5 (it is also corollary for planner graph ) it means minimum degree can not be 6 , minimum degree will be <=5
therefore option (d) will be the answer