see its very simple

 

the equation given is:-     A + 2B ----------> 2C

 so the initial concentrations of   A = 2mols

                                               2B= 2*3 mols

                                              2 C =2*0.2mols

 

AT equlib given concentration of C would be 2* 0.5 mols

so at equlibrium 0.6 mols of C are formed frm 0.3 mols of A n 0.6 moles of B

                 

       conc of A=2-0.3=1.7

       conc of B=6-0.6=5.4

 

nw , Kc= [C]²/[A][B]²

 

Kc=   1/(1.7)(5.4)²

 

Hence Kc =0.026(approx.)           

 

ans is b)