You're starting with
(12.0 mL) (0.550 M) = (0.0120 L) (0.550 M) = 0.00660 mol
of FeCl₃.
For every mole of FeCl₃ in the balanced reaction, you need 3 moles of NH₄OH, so you need a minimum 0.0198 mol NH₄OH per mole FeCl₃.
Then the requisite volume of NH₄OH solution needed is V such that
(0.0198 mol) / V = 0.200 M
Solve for V to get
V = (0.0198 mol) / (0.200 M) = 0.0990 L = 99 mL
