As you know, the equilibrium constant for a given chemical equilibrium depends on
More specifically, the equilibrium constant is defined as the ratio between the product of the equilibrium concentrations of the products and the product of the equilibrium concentrations of the reactants, each raised to the power of their stoichiometric coefficients.
Your equilibrium looks like this
$"N"_text(2(g]) + color(red)(3)"H"_text(2(g]) rightleftharpoons color(blue)(2)"NH"_text(3(g])$
By definition, the equilibrium constant for this reaction,
$K_c = (["NH"_3]^color(blue)(2))/(["N"_2] * ["H"_2]^color(red)(3))" "$ , where
In your case, you know that these concentrations are equal to
$["NH"_3] = "3.0 M"$
$["H"_2] = "1.0 M"$
$["N"_2] = "2.0 M"$
This means that the equilibrium constant will be equal to - keep in mind that equilibrium constant are usually used without corresponding units
$K_c = 3.0^color(blue)(2)/(2.0 * 1.0^color(red)(3)) = 9/2 = color(green)(4.5)$
If you want, you can include the units in the calculation to get
$K_c = (3.0^color(blue)(2) "M"^color(blue)(2))/("2.0 M" * 1.0^color(red)(3)"M"^color(red)(3)) = "4.5 M"^(-2)$