Gravimetric analysis
The amount of $\ce{Mg2P2O7}$ in 1.054 gram of a pure sample is:
$$ n = \frac{\pu{1.054g}}{\pu{222.57g mol-1}} = \pu{0.004736 mol}$$

There are two atoms of $\ce{P}$ in $\ce{Mg2P2O7}$, but only one in $\ce{NaH2PO4}$. Without considering the two reactions done in detail, we can infer a 2:1 ratio:

$$\ce{2 NaH2PO4 + ... -&gt; Mg2P2O7 + ...}$$

Thus, the mass of $\ce{NaH2PO4}$ is
$$  m_\ce{NaH2PO4} = 2 \cdot \pu{0.004736 mol} \cdot \pu{120.0 g mol-1}= \pu{1.137g} $$