So, you know that you're dealing with an equilibrium reaction that has its equal to 96.2 at a certain temperature.
The fact that $K_c$ is larger than one tells you that the reaction will favor the product, $PCl_5$, at equilibrium. Moreover, since you only start with reactants, you can predict that the concentrations of both $PCl_3$, and of $Cl_2$ will decrease.
At the same time, the concentration of $PCl_5$ will increase. Use an ICE table to help you determine the equilibrium concentrations for your reaction
$" "PCl_(3(s)) + Cl_(2(g)) rightleftharpoons PCl_(5(g))$
I.....0.22...........0.42.................0
C.....(-x).............(-x)..................(+x)
E...0.22-x.......0.42-x................x
By definition, the equilibrium constant will be equal to
$K_c = ([PCl_5])/([PCl_3] * [Cl_2]) = x/((0.22-x) * (0.42 - x)) = 96.2$
Rearrange this equation to quadratic form
$96.2 * (0.22-x) * (0.42-x) = x$
$96.2 * (0.0924 - 0.22x - 0.42x + x^2) = x$
$96.2x^2 - 62.568x + 8.888 = 0$
This equation will produce two for $x$
${ (cancel(x_1 = 0.4408)), (x_2 = 0.2096) :}$
The first solution is eliminated because it will result in negative equilibrium concentrations for $PCl_3$ and $Cl_2$, which means that you'll get
$[PCl_3] = 0.22 - 0.2096 = color(green)("0.0104 M")$
$[Cl_2] = 0.42 - 0.2096 = color(green)("0.210 M")$
$[PCl_5] = 0 + 0.2096 = color(green)("0.210 M")$
SIDE NOTE I've left the answers with three .
