Calculating weight of soluble salt in a solution
Ok... It is hard to read you post since the details are scattered around.  

Problem Statement

Given: 50 ml of a 0.250 molar solution of barium nitrate with a density of 1.035 g/ml  and that barium phosphate has a $K_\mathrm{sp} = 4.03 \times 10^{-25}$. 

Question: How much $\ce{Na3PO4}$ must be added to reduce the concentration of barium to less than 7.00 ppb.  

Complications with problem

Initial density


  The density of the initial solution, 1.035 g/ml, is useless. 


Final density


  We need a density for the final solution to convert 7.00 ppb $\ce{Ba^{2+}}$ to a concentration of grams per liter. There is no density for the final solution, so the assumption will be that the density is 1.000 g/ml which means that a liter of the final solution will weigh 1000. grams. Since each $\ce{Ba^{2+}}$ cation will be replaced by 2 $\ce{Na^+}$ cations, the final solution will be at least 0.50 molar in $\ce{NaNO3}$. Thus the assumed density of 1.000 g/ml will be too low. 


A pragmatic point here about the experimental procedure.


  You typically wouldn't want to add solid $\ce{Na3PO4}$ to a solution containing $\ce{Ba^{2+}}$ since $\ce{Ba3(PO4)2}$ would precipitate on the particles of $\ce{Na3PO4}$ sealing them from the solution and thus preventing complete dissolving of the $\ce{Na3PO4}$ particles. So experimentally you first need to dissolve the $\ce{Na3PO4}$ in a small amount of water and then add the solution of $\ce{Na3PO4}$ to the solution of $\ce{Ba(NO3)2})$. Thus the initial volume would really be over 50 ml.  


Rounding errors


  The calculations will be carried to 5 significant figures to try to avoid rounding errors due to intermediate results. The final result will be properly rounded to 3 significant figures


Calculations

Moles of barium in the initial solution

In the initial solution $\ce{[Ba]} = 0.250\ \mathrm{m}$ so 50 ml contains

$\mathrm{m}\ \ce{Ba^{2+}} = 0.250\ \mathrm{m/l}\times 0.050\ \mathrm{l} = 0.0125\ \ \mathrm{m}$

Moles of barium in the final solution

In the final solution for Ba:

$\ce{Ba^{2+}} = 7.00\ \mathrm{ppb}$
$\quad\quad = (7.00\times 10^{-9})\times 1000\ \mathrm{g/l}=7.00\times 10^{-6}\ \mathrm{g/l}$
$\quad\quad  = \dfrac{7.00\times 10^{-6}\ \mathrm{g/l}}{137.33\ \mathrm{ g/m}} = 5.0972\times 10^{-8}\ \mathrm{m/l} $

For 50 ml of solution this is:
$\mathrm{m}\ \ce{Ba^{2+}} = 5.0972\times 10^{-8}\ \mathrm{m/l}\times 0.050\ 
\mathrm{l} = 2.5486\times10^{-9}\ \ \mathrm{m}$

$2.5486\times10^{-9}\ \ \mathrm{m} \ll 0.0125\ \ \mathrm{m}$ so the residual amount of barium in solution can be ignored.


  
    (1) *In determining the amount of barium phosphate that will precipitate, the amount of barium left in solution is so small that it doesn't matter if the solution has a density of 1.000 g/ml or 1.05 g/ml. So the density of the sodium nitrate solution doesn't really matter for that part of the problem.
    
    (2) $2.5486\times10^{-9}\ \ \mathrm{m} \ll 0.0125\ \ \mathrm{m}$ This is a significant figures argument. In other words for intermediate calculations I'm using only 5 significant figures.
    $$\begin{align}
0&amp;.012\,500\,000\,000\,0 \\
-0&amp;.000\,000\,002\,548\,6\\
\hline
0&amp;.012\,499\,998\,551\,4 
\end{align}$$ Now when trimmed to 5 significant figures the intermediate result is 0.012500. But the 0.0125 is exact, so I don't need the two extra zeros at the end. 
  


Residual phosphate in the final solution

For the residual phosphate, in excess of that needed to precipitate the stoichometric amount of barium is:
$K_\mathrm{sp} = \ce{[Ba^{2+}]^3[PO4^{3-}]^2}$ or $\ce{[PO4^{3-}]} = \sqrt{\dfrac{K_\mathrm{sp}}
{[Ba^{2+}]^3}}$

$\ce{[PO4^{3-}]} = \sqrt{\dfrac{4.03\times10^{-25}}{(5.0972\times 10^{-8})^3}} = 5.5164\times10^{-2}\ \mathrm{m/l}$

For 50 ml this would be:

$\mathrm{g}\ \ce{PO4^{-3}}\ \mathrm{as }\ \ce{Na3PO4} = (5.5164\times10^{-2}\ \mathrm{m/l})\times 0.050\ \mathrm{l}\times 163.94\ \mathrm{g/m} = 0.45218\ \mathrm{g}$  


  This is too great of an amount of residual phosphate in solution to be ignored. Thus we need to add this to how many grams of trisodium phosphate are needed to precipitate 0.0125 moles of $\ce{Ba^{2+}}$ as $\ce{Ba3(PO4)2}$. 


Stoichiometric amount of phosphate to precipitate all the barium

Each mole of $\ce{Ba3(PO4)2}$ requires 3 moles of $\ce{Ba^{2+}}$. So for 0.0125 moles of $\ce{Ba^{2+}}$ there would be

$\mathrm{moles}\ \ce{Ba3(PO4)2} = \dfrac{\mathrm{moles}\ \ce{Ba^2+}}{3}= \dfrac{0.0125}{3} = 4.1667\times10^{-3}$

But each mole of $\ce{Ba3(PO4)2}$ requires two moles of $\ce{PO4^{3-}}$

$\mathrm{moles}\ \ce{PO4^{3-}} = 2\times(\mathrm{moles}\ \ce{Ba3(PO4)2})= 2\times( 4.1667\times10^{-3}) = 8.3333\times10^{-3}$

$\ce{Na_3PO4} = \mathrm{MW}\ 163.94\ \mathrm{g/m}\\ 
\quad\quad\quad\quad= 163.94\times8.3333\times10^{-3} = 1.3662\ \mathrm{g}$

Total amount of trisodium phosphate needed

$\mathrm{Total}\ \ce{Na_3PO4} = 1.3662 + 0.4522 = 1.8184\ \mathrm{g}$

Now rounding the final result to 3 significant figures we get 1.82 grams.

Now the other pertinent fact is how much $\ce{Na3PO4}$ can be dissolved in 50 ml at 25 °C? Wikipedia shows 14.5 g/100 mL (25 °C), so 7.25 g/50 mL (25 °C). Thus it should be possible to dissolve the $\ce{Na3PO4}$.