Phosphate buffer from phosphoric acid and K2HPO4?
I suggest you first consider the following reaction:
$$\ce{H3PO4 + 3K2HPO4 -&gt; 2HPO4^{2-} + 2H2PO4- + 6K+}$$

It appears, that transforming all $\ce{H3PO4}$ to equal amounts of $\ce{HPO2-}$ and $\ce{H2PO4-}$ 
requires 3 mole equivalents of $\ce{K2HPO4}$. If you add 3 mole equivalents of $\ce{K2HPO4}$ you will end up in a situation where the concentration of $\ce{[HPO2^{-}] = [H2PO4^{-}]}$, i.e. at the $\ce{pH} = pK_{a2} = 7.21$. Then, I suppose you use the $\ce{HH}$-equation to figure out the rest. 

At pH = pka2 = 7.21 the concentration of [H2PO4(-)] = [HPO4(2-)] = 0.40 M. This is because we have added 3 mole equivalents of K2HPO4 to 50*0.2 = 10 mmole of phosphoric acid, i.e. we have reached a total concentration of phosphoric acid protolytes of (3*50*0.2 + 50*0.2)/50 = 0.80 M. 

Now, since we wanted to reach pH = 7.0, we have theoretically added too much of K2HPO4. We suppose the excess amount is equal to x. 

At pH = 7.0: [HPO4(2-)] &lt; [H2PO4(-)].   At  this pH, only HPO4(2-) and H2PO4(-) are present in significant amounts in the solution. We can then calculate the following:
7.00  = 7.21 + log ([HPO4(2-)] - x/[H2PO4(-)]) = 7.21 + log (0.4 - x)/0.4) => x = 0,1533.

Now, initially we had 50*0.2 mmole of phosphoric acid. To reach pH = 7.0 we should then add 3*50*0.2 - 0.1533*50 mmole =  30 - 7,66(5) = 22,34 mmole of K2HPO4 or 3.8(9) gram.

Please check the calculations.