Answer:
Kc = 8.05x10⁻³
Explanation:
This is the equilibrium:
                  2NH₃(g)   ⇄     N₂(g)     +     3H₂(g) 
Initially       0.0733
React         0.0733α          α/2                3/2α
Eq     0.0733 - 0.0733α    α/2                0.103
We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.
Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.
3/2α = 0.103
α = 0.103 . 2/3 ⇒ 0.0686
So, concentration in equilibrium are
NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682
N₂ = 0.0686/2 = 0.0343 
So this moles, are in a volume of 1L, so they are molar concentrations.
Let's make Kc expression:
Kc= [N₂] . [H₂]³ / [NH₃]² 
Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³