Let the given reactions be: (i) $\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(s) + 3\text{CO}_2(g); \Delta H_1 = -26.88\text{ kJ}$ (ii) $\text{FeO}(s) + \text{CO}(g) \rightarrow \text{Fe}(s) + \text{CO}_2(g); \Delta H_2 = -16.5\text{ kJ}$

The target reaction is: $\text{Fe}_2\text{O}_3(s) + \text{CO}(g) \rightarrow 2\text{FeO}(s) + \text{CO}_2(g); \Delta H = ?$

By applying Hess's Law, we can obtain the target reaction by reversing equation (ii), multiplying it by 2, and then adding it to equation (i): Reverse equation (ii) and multiply by 2: $2\text{Fe}(s) + 2\text{CO}_2(g) \rightarrow 2\text{FeO}(s) + 2\text{CO}(g); \Delta H_3 = 2 \times (+16.5\text{ kJ}) = +33.0\text{ kJ}$

Now, add this to equation (i): $[\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g)] + [2\text{Fe}(s) + 2\text{CO}_2(g)] \rightarrow [2\text{Fe}(s) + 3\text{CO}_2(g)] + [2\text{FeO}(s) + 2\text{CO}(g)]$ $\Rightarrow \text{Fe}_2\text{O}_3(s) + \text{CO}(g) \rightarrow 2\text{FeO}(s) + \text{CO}_2(g)$

Therefore, the enthalpy change for the target reaction is: $\Delta H = \Delta H_1 + \Delta H_3 = -26.88\text{ kJ} + 33.0\text{ kJ} = +6.12\text{ kJ} \approx +6.2\text{ kJ}$