First consider what happened when the HNO3 was added to the KOH.
The equation for that reaction is: HNO3 + KOH ---> KNO3 + H2O. So 1 mole of HNO3 neutralizes 1 mole of KOH.
In the original mixture there were 100 mL = .1 L of .160 M HNO3, for a total of .016 moles of HNO3.
And there were 400 mL = .4 L of .1 M KOH, for a total of .04 moles of KOH.
So .016 moles of KOH were neutralized, leaving .024 moles still unneutralized.
Now look at the equation when HCl is added: HCl + KOH ---> KCl + H2O.
So 1 mole of HCl neutralizes 1 mole of KOH. Since the there are .024 moles of KOH in the mixture, .024 moles of HCl are needed to neurtralize it. If the HCL solution is 2.0 M, that means that .012 L = 12 mL of the HCl solution must be added to neutralize the mixture.