1. Write the equilibrium reaction

$\mathrm{Fe_2O_{3(s)} + 3CO_{(g)} \rightleftharpoons Fe_{(l)} + 3CO_{2(g)}}$

We are asked: which change will not disturb the equilibrium according to Le-Chatelier's principle.

2. Important equilibrium idea

For a heterogeneous equilibrium, the activities of pure solids and pure liquids are constant, so they do not appear in the equilibrium expression.

Thus, for this reaction,

$K = \frac{(P_{CO_2})^3}{(P_{CO})^3}$

Here, $\mathrm{Fe_2O_3(s)}$ and $\mathrm{Fe(l)}$ do not appear in $K$.

3. Check each option

Option A: Addition of $\mathrm{CO_2}$

Adding a product increases $P_{CO_2}$, so the reaction quotient increases. The system will shift left to oppose this change. So equilibrium is disturbed.

Option B: Removal of $\mathrm{CO}$

Removing a reactant decreases $P_{CO}$, changing the reaction quotient. The system will shift left to produce more $\mathrm{CO}$. So equilibrium is disturbed.

Option C: Addition of $\mathrm{Fe_2O_3}$

$\mathrm{Fe_2O_3}$ is a solid. Changing the amount of a pure solid does not affect the equilibrium position, as long as some solid is already present. So equilibrium is not disturbed.

Option D: Removal of $\mathrm{CO_2}$

Removing a product decreases $P_{CO_2}$. The system will shift right to form more $\mathrm{CO_2}$. So equilibrium is disturbed.

4. Conclusion

The only change that does not disturb the equilibrium is:

$\boxed{\text{C: Addition of } \mathrm{Fe_2O_3}}$