Answer
75.0 mL of 0.150 M $Na_3PO_4$ has the most ions
Work Step by Step
$NaOH: 0.1000*0.100*6.0221*10^{23}*2=1.2*10^{22}$ ions
$BaCl_2: 0.0500*0.200*6.0221*10^{23}*3=1.81*10^{22}$ ions
$Na_3PO_4: 0.0750*0.150*6.0221*10^{23}*4=2.71*10^{22}$ ions