Every time you have to solve problems in which you are asked to determine how much of something is needed to react with a given mass of something else, you must think moles and, more specifically, mole ratios.
Start with the balanced chemical equation
#2KI(aq) + Pb(NO_3)_2(aq) -> PbI_2(s) + 2KNO_3(aq)#
This is a double replacement reaction that leads to the formation of an insoluble precipitate,
Notice the mole ratio between
Use the molar masses of the compounds to go from grams to moles (for
#55.2 cancel("g") * ("1 mole")/(331.2 cancel("g")) = "0.167 moles"# #Pb(NO_3)_2#
This means that you need
#0.167 cancel("moles"color(white)(a) Pb(NO_3)_2) * ("2 moles " KI)/(1 cancel("mole"color(white)(a) Pb(NO_3)_2)) = "0.334 moles"color(white)(a) KI#
The mass will be
#0.334 cancel("moles") * ("166 g")/(1 cancel("mole")) = "55.4 g"color(white)(a) KI#
As a conclusion, always go to moles, it'll make your life easier, and remember to use mole ratios.
Here's a video of the reaction, if you're curios about what the precipitate looks like: