Since you’ve got the first part I’ll just answer the second question so to speak.
So you’ve got your balanced chemical equation for this reaction
##AgNO_(3(aq)) + KCl_((aq)) -> AgCl_((s)) + KNO_(3(aq))##
You know that you have a ##1:1## between all the species involved. This means that the number of moles of potassium chloride must be equal to the number of moles of silver nitrate.
Since I assume you’ve calculated the number of moles of silver nitrate to be
##C = n/V => n = C * V##
##n_(AgNO_3) = 0.162 M * 1.27 L = 0.206 moles## ##AgNO_3##
automatically you’ll have
##0.206cancel(moles AgNO_3) * (1 mole KCl)/(cancel(1 mole AgNO_3)) = 0.206 moles## ##KCl##
Now use the volume of potassium chloride given to figure out what the solution’s must be in order to have that many moles of ##KCl## available for the reaction
##C = n/V = 0.206 moles KCl/3.78 L = color(red)(0.0545 M)##