Answer:
Approximately [tex]\rm -89 \; kJ\cdot mol^{-1}[/tex].
Assumption: the density of the solution is equal to the density of pure water.
Explanation:
The enthalpy of neutralization is defined as the enthalpy change for each moles of water produced. (Clark, Physical & Theoretical Chemistry, Chemistry Libretexts.)
Each mole of [tex]\rm NaOH[/tex] formula units will neutralize one mole of [tex]\rm HCl[/tex] to produce one mole of water. [tex]\rm HCl[/tex] and [tex]\rm NaOH[/tex] are available at equal volume and concentration. In other words, there's an equal number of both reactants. All [tex]\rm HCl[/tex] and [tex]\rm NaOH[/tex] will react to form water.
[tex]V(\mathrm{HCl}) = \rm 137\; cm^{3} = 0.137\;dm^{3}[/tex].
[tex]V(\mathrm{NaOH}) = \rm 137\; cm^{3} = 0.137\;dm^{3}[/tex].
[tex]n = c\cdot V = \rm 0.137\;dm^{3} \times 2.6\;mol\cdot dm^{-3} = 0.3652\; mol[/tex].
In other words, there are [tex]\rm 0.3652\; mol[/tex] of [tex]\rm HCl[/tex] and [tex]\rm NaOH[/tex] each. The two will react to produce [tex]\rm 0.3652\; mol[/tex] of water.
How much heat is released?
Assume that the volume of the liquid is equal to the volume of the [tex]\rm HCl[/tex] solution plus the volume of the [tex]\rm NaOH[/tex] solution. That's [tex]\rm 0.274\;dm^{3}[/tex]. Assume that the density of the solution is equal to that of water under room temperature. [tex]\rho(\text{water}) = \rm 1.000\; kg\cdot dm^{-3}[/tex]. The mass of the liquid will be [tex]m = \rho \cdot V = \rm 0.274\; dm^{3} \times 1.000\; kg\cdot dm^{-3} = 0.274\; kg = 274\;g[/tex].
Change in temperature:
[tex]\Delta T = \rm 325.8 - 298 = 27.8\; K[/tex].
Heat that the solution absorbed:
[tex]Q = c\cdot m \cdot \Delta T = \rm 4.18\;J\cdot K^{-1}\cdot g^{-1} \times 274\; g\times 27.8\;K = 36410.216\; J = 36.410216\; kJ[/tex].
That will also be the amount of heat released from the reaction if there's no energy loss.
[tex]\displaystyle \Delta H(\text{Neutralization}) = \frac{-Q}{n(\text{water produced})} = \rm \frac{36.410216\; kJ}{0.3652\; mol} \approx 89\; kJ\cdot mol^{-1}[/tex].
