How do I calculate the enthalpy change when a gas is being used to heat water?
You're on the right track.

\begin{eqnarray}
Q &amp; = &amp; m\cdot c\cdot \Delta T\\
\mathbf{unit\ analysis} \qquad \textrm{kJ} &amp; = &amp; \textrm{kg}\cdot \frac{\textrm{kJ}}{\textrm{kg} \cdot \textrm{K}} \cdot \textrm{K}
\end{eqnarray}

For water, the specific heat capacity $c = 4.18 \large{\frac{\textrm{kJ}}{\textrm{kg}\cdot \textrm{K}}}$. You have that right.

The temperature difference $\Delta T = (50.7 - 7.44) = 43.26\ \textrm{K}$. We're talking about differences, so we can use kelvin as the unit. You have that right too.

Now, what is the mass of $724\ \textrm{cm}^3$ water? 

Choose any unit you want, but adjust the outcome of your calculation according to the unit analysis above!

Now you know the kJ you need.

From the combustion equation, you know the amount of heat delivered by $n$ moles of butane. Note the $n$!

Now you can calculate how many moles of butane you will need.

From there, the calculation of the mass is not difficult either.