The equilibrium constant for this reaction will be equal to 3.13.
So, you know that you're dealing with a generic equilibrium that involves three species, #A#, #B#, both reactants, and #C#, the product.
Even before doing any calculations, you can predict that #K_c# will be bigger than 1. Notice that the concentration of #A# decreases, while the concentration of #C# increases.
This tells you that the initial concentrations will cause the equilibrium to shift right, i.e. the reaction quotient, #Q_c#, is smaller than #K_c#.
If this is the case, the equilibrium concentration of #B# will decrease as well, but not by the same amount as the concentration of #A# - this happens because of the difference between the stoichiometric coefficients of the two reactants.
So, you were only given two equilibrium concentrations, more specifically those of #A# and of #C#. However, you can use the initial concentrations to determine what the equilibrium concentration of #B# would be.
To do that, use an ICE table for you equilibrium reaction
#" " " "A " "+ " "color(red)(2)B " " rightleftharpoons " "C#
I......0.650............1.20...............0.700
C.......(-x)..............(-#color(red)(2)#x)...................(+x)
E.....0.650-x........1.20-2x..........700+x
You know that #[A]_"equilibrium" = "0.450 M"#. This means that
#0.650 - x = 0.540 => x = 0.200#
Check to see if it matches the equilibrium concentration of #C#
#[C]_"equilibrium" = 0700 + x = 0.700 + 0.200 = "0.900 M"#
This means that the equilibrium concentration of #B# will be
#[B]_"equilibrium" = 1.2 - 2x = 1.2 - 2 * 0.200 = "0.800 M"#
Therefore, the equilibrium constant for this reaction will be
#K_c = ([C])/([A] * [B]^color(red)(2)) = 0.900/(0.450 * 0.800"^2) = color(green)(3.13)#
SIDE NOTE The value of #Q_c# at the start of the reaction was 0.74, which again confirms our initial prediction.
