If 3.8 g of Cu3(PO4)2 was recovered from step 1

In mL, how much of 3Cu(NO3)2 and 2Na3PO4 will I need to make 10 grams of Cu3(PO4)2?
Balanced equation: 2Na3PO4(ac) + 3Cu(NO3)2(ac) = Cu3(PO4)2(s) + 6NaNO3(ac)

Na3PO4 molarity: (1.0 M)

STEP 1. , Find the molar mass of the following (use a Periodic Table):

Cu3(PO4)2 =3(63.55) + 2(31 +64) = 380.65

STEP 2.  Change the grams to moles.

moles = grams/ molar mass

10 g Cu3(PO4)2 = 10/380.65 = 0.0263 mol

STEP 3. Use the mole ratios.

By the mole ratios (look at the coefficients from the equation), it requires 3 moles Cu(NO3)2 and 2 moles Na3PO4 to form 1 mole of Cu3(PO4)2

This means that to produce 0.0263 mole Cu3(PO4)2, 

3 x 0.0263 = 0.0789 mole Cu(NO3)2 and

2 x 0.0263 = 0.0526 mol Na3PO4 are required.

STEP 4. Use the Molarity formula to find the volume.

or Liters = moles/ Molarity

Volume of Cu(NO3)2 = 0.0789 mol/ 5.0 M = 0.01578 L ≈ 16 mL

Volume of Na3PO4 = 0.0526 mol/ 1.0 M = 0.0526 L ≈ 53 mL

I see you have the molarity for the two solutions but you would need a bit more information like the starting mass from one of the products or liters of solution.