The chemical equation is

#"N"_2 + "3H"_2 ⇌ "2NH"_3#

The equilibrium moles of each component are:

#"N"_2 = "0.1 mol"#

#"H"_2 = "0.2 mol"#

#"NH"_3 = "(2 - 0.1 - 0.2) mol" = "1.7 mol"#

The volume of the container is #"0.500 dm"^3#, so the equilibrium concentrations of each component are

#["N"_2] = "0.1 mol"/"0.500 dm"^3 = "0.20 mol/dm"^3#

#["H"_2] = "0.2 mol"/"0.500 dm"^3 = "0.40 mol/dm"^3#

#["NH"_3] = "1.7 mol"/"0.500 dm"^3 = "3.4 mol/dm"^3#

We can set up part of an ICE table to solve this problem.

#color(white)(mmmmmmm)"N"_2 + "3H"_2 ⇌ "2NH"_3#

#"E/mol·dm"^"-3": color(white)(l)"0.20color(white)(ml)0.40color(white)(mml)3.4#

The equilibrium constant expression is

#K_text(c) = (["NH"_3]^2)/(["N"_2]["H"_2]^3)#

∴ #K_text(c) = 3.4^2/(0.20 × 0.40^3) = 900#