The idea here is that you need to write a balanced chemical equation for the dissociation of chlorine gas,
So, you will have
$"Cl"_ (2(g)) rightleftharpoons color(red)(2)"Cl"_ ((g))$
Notice the every mole of chlorine gas that dissociates produces
You know that the reaction vessel initially contained
$6 color(red)(cancel(color(black)("moles Cl"_2))) * ("50 moles Cl"_2)/(100color(red)(cancel(color(black)("moles Cl"_2)))) = "3 moles Cl"_2$
You can thus say that at equilibrium, the reaction vessel will contain
$n_(Cl_2) = "6 moles" - overbrace("3 moles")^(color(blue)("50% of the initial amount")) = "3 moles Cl"_2$
and
$n_(Cl) = "0 moles" + color(red)(2) xx "3 moles" = "6 moles Cl"$
Use the volume of the vessel to calculate the concentrations of the two species
$["Cl"_2] = "3 moles"/"3 L" = "1 M"$
$["Cl"] = "6 moles"/"3 L" = "2 M"$
By definition, the *equilibrium constant for this reaction,
$K_c = (["Cl"]^color(red)(2))/(["Cl"_2])$
Plug in your values to find
$K_c = ("2 M")^color(red)(2)/"1 M" = ("4 M"^color(red)(cancel(color(black)(2))))/(1color(red)(cancel(color(black)("M")))) = "4 M"$
The equilibrium constant is usually given without added units, but keep in mind that, as you can see in this example, that is not always the case
$K_c = color(green)(|bar(ul(color(white)(a/a)color(black)(4)color(white)(a/a)|)))$
The answer is rounded to one , the number of sig figs you have for the number of moles of chlorine gas added to the reaction vessel and for the volume of the vessel.