According to the given condition:

$$T_1= 20.2^0 C$$

$$T_2=26.3 ^0 C$$

$$C_{CuSO_4}=50\ ml$$

$$M_{CuSO_4} =1.00\ mol/ml$$

$$V_{KOH}=2\ M$$

The heat change $$(\Delta H)$$ of the reaction is given as:

$$\Delta H =-C_pm_c\Delta T$$

Where:

$$C_p= specific \ heat \ constant\ of \ water$$

The specific heat constant of water is $$4.184 J/g^0 C$$

$$m_c=mass\ of\ calorimeter\ solution$$

The density of water is $$1g/ml$$

Therefore mass of the solution=volume$$= 50+50=100g$$

So, $$m_c=100g$$

$$\Delta T=T_2-T_1= 26.3-20.2=6.1 ^0C$$

Substituting these values in the above expression, we get

$$\Delta H=-(4.184\ J/g^0 C) \ (100g)( 6.1^0 C)$$

$$=-2567.94\ J$$

$$\therefore \Delta H= -2.57\ kJ$$

Hence the enthalpy of reaction is -2.57 kJ