Using Hess's law to find molar enthalpy of formation without an initial reaction
Since you are asked to find the enthalpy of formation for ammonia, it's convenient to write the equation of ammonia synthesis normalized for $\pu{1 mol}$ of $\ce{NH3}$:

$$\ce{0.5 N2 + 1.5 H2 -&gt; NH3}$$

To obtain this equation and apply Hess's law, the following linear combination of the provided equations should be used:

$$
\begin{align}
\ce{4 NH3 + 3 O2 &amp;-&gt;6 H2O + 2N2} &amp; ΔH_1 &amp;= \pu{-1516 kJ} &amp;&amp;|\cdot(-0.25) \tag{1} \\
\ce{2 H2 + O2 &amp;-&gt; 2 H2O} &amp; ΔH_2 &amp;= \pu{-572 kJ} &amp;&amp;|\cdot 0.75 \tag{2} \\
\hline
\ce{0.5 N2 + 1.5 H2 &amp;-&gt; NH3} &amp; ΔH_3 &amp;= -0.25ΔH_1 + 0.75ΔH_2 \tag{3}
\end{align}
$$

So the final answer is

$$ΔH_3 = -0.25\cdot(\pu{-1516 kJ}) + 0.75\cdot(\pu{-572 kJ}) = \pu{-50 kJ}$$

The catch here is that you are asked to find molar enthalpy of formation, and calculating for the "convenient" equation with integer coefficients

$$\ce{N2 + 3 H2 -&gt; 2 NH3}$$

you eventually would obtain the doubled value of $\pu{-100 kJ mol-1}$ referred to $\pu{2 mol}$ of ammonia (I almost fell for it, too).