Calculating the heat of reaction for Mg metal and HCl
The equations you listed can be used to partially solve the problem.  However, with the equations you used, you have made one error which was in your choice of "moles."  Your moles value must match up with your $\ce{C_{p}}$, so you must find the moles of Water in $\pu{95.00mL}$!  Alternatively, you can use $4.184 \pu{\frac{J}{g\ ^\circ C}}$ and $\pu{1 \frac{g}{mL}}$ and $\pu{95.00mL}$ of water along with the equation $\mathrm{Q = n \times C_p \times \Delta T}$.

On the other hand, the premise of your solution is close but wrong, which is obvious if you examine the units:


  $\pu{0.1773 mol} \times 75.3 \frac{\pu J}{\pu{mol K}} \times \pu{9.2 K} = \pu{122.8 J}$


The units for Heat of Reaction are $\pu{\frac{J}{mol}}$, not $\pu{J}$!

To finish off your solution, you must divide by the moles of reaction that occurred (given by the limiting reactant).  $\pu{-44.4 kJ}$ is the value calculated based on the moles of Mg reacted.  However, I suspect that the given answer is wrong because the balanced equation does not yield a 1:1 reaction ratio between $\ce{Mg}$ and $\ce{HCl}$, which means that $\ce{HCl}$ is the limiting reactant, which would make the Heat of Reaction Larger.