The answer is #1.2M#.

First, start with the formula equation

#Pb(NO_3)_2(aq) + 2KCl(aq) -> PbCl_2(s) + 2KNO_3(aq)#

The complete ionic equation is

#Pb^(2+)(aq) + 2NO_3^(-)(aq) + 2K^(+)(aq) + 2Cl^(-)(aq) -> PbCl_2(s) + 2K^(+)(aq) + 2NO_3^(-)(aq)#

The net ionic equation, obtained after eliminating spectator ions (ions that can be found both on the reactants, and on the products' side), is

#Pb^(2+)(aq) + 2Cl^(-)(aq) -> PbCl_2(s)#

According to the solubility rules, lead (II) chloride can be considered to be insoluble in water.

Notice that we have a #1:2# mole ratio between #Pb(NO_2)_2# and #KCl# in the formula reaction; this means that 2 moles of the latter are needed for every 1 mole of the former for a complete reaction.

We know that #Pb(NO_3)_2#'s molarity, which is defined as the number of moles of solute divided by the liters of solution, is #0.400M#. THis means that

#n_(Pb(NO_3)_2) = C * V = 0.400M * 30.0 * 10^(-3)L = 0.012# moles

The number of #KCl# moles will then be

#n_(KCl) = 2 * n_(Pb(NO_3)_2) = 2 * 0.012 = 0.024# moles

This makes the potassium chloride's molarity equal to

#C_(KCl) = n/V = (0.024 mol es)/(20.0 * 10^(-3)L) = 1.2M#