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I have 0.400*0.13=5.2 so far

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What volume of 0.300 M Na3PO4 is required to precipitate all the lead(II) ions from 130.0 mL of 0.400 M Pb(NO3)2?

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We would do well to stay on the same problem and repost over and over.
First, 0.400 x 0.13 = 0.052
Step 2. Convert mols Pb(NO3)2 to mols Na3PO4 using the coefficients in the balanced equation.
Step 3. Then M Na3PO4 = mols Na3PO4/L Na3PO4. You have mols and M, solve for L.