`Na_3PO_4` will ionized as follows in an aqueous solution.
`Na_3PO_4 rarr 3Na^++PO_4^(3-)`
As we can see from the balanced equation one `Na_3PO_4` will give us three `Na^+` moles.
Amount of `Na_3PO_4 = 0.4/1000xx20 = 0.008mol`
`Na_3PO_4:Na^+ = 1:3`
Amount of `Na^+` `= 0.008xx3 = 0.024mols`
So we have 0.024 moles of `Na^+` ions.