The pKa value is given wrong. The actual value of pKa3 = 12.35
According to Henderson-Hasselbulch equation:
pH = pKa + Log[Na3PO4.H2O/Na2HPO4.2H2O]
12.5 = 12.35 + Log[Na3PO4.H2O/Na2HPO4.2H2O]
i.e. Log[Na3PO4.H2O/Na2HPO4.2H2O] = 0.15
i.e. [Na3PO4.H2O/Na2HPO4.2H2O] = 100.15 = 1.4125 ---- Equation 1
[Na2HPO4.2H2O + Na3PO4.H2O] = 250 mL * 0.2 mmol/mL = 50 mmol ---- Equation 2
From equations 1 and 2:
[Na2HPO4.2H2O] = 50/(1.4125+1) = 20.725 mmol
Now, [Na3PO4.H2O] = 50 - 20.725 = 29.275 mmol
The mass of Na2HPO4.2H2O required to prepare the given buffer = 20.725 mmol * 178 g/mol= 3.689 g
The mass of Na3PO4.H2O required to prepare the given buffer = 29.275 mmol * 182 g/mol = 5.328 g
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