What volume of 0.205 m k3po4 solution is necessary to completely react with 154 ml of 0.0110 m nicl2? 1.65 l?

The balanced equation for the above reaction is 
2K₃PO₄   + 3NiCl₂  ---> 6KCl + Ni₃(PO₄)₂
stoichiometry of K₃PO₄  to NiCl₂ is 2:3
the number of NiCl₂ moles reacted - 0.0110 mol/L x 0.154 L = 1.69 x 10⁻³ mol
if 3 mol of NiCl₂ reacts with - 2 mol of K₃PO₄ 
then 1.69 x 10⁻³ mol of NiCl₂ reacts with - 2/3 x 1.69 x 10⁻³  = 1.13 x 10⁻³ mol of K₃PO₄
molarity of K₃PO₄ solution given - 0.205 M
there are 0.205 mol in 1 L
therefore 1.13 x 10⁻³ mol are in - 1.13 x 10⁻³ mol / 0.205 mol/L = 5.51 mL
volume of K₃PO₄ required - 5.51 mL

Ibelieve the answer is d