Using the omitted variable formula, we know that:

$$\hat{\beta}_1 = \beta_1 + \beta_2 \delta$$

Where:

$$\delta = \frac{\text{Cov}(X_1, X_2)}{\text{Var}(X_1)} = \frac{\rho_{X_1X_2} \sigma_{X_1} \sigma_{X_2}}{\text{Var}(X_1)}$$

Given:

• $\hat{\beta}_1 = 2.4$
• $\beta_1 = 2.2$
• $\beta_2 = 1.1$
• $\sigma^2_{X_1} = 16 \Rightarrow \sigma_{X_1} = 4$
• $\sigma^2_{X_2} = 49 \Rightarrow \sigma_{X_2} = 7$

Substitute into the formula: $`$2.4` = 2.2 + 1.1 \cdot \delta \Rightarrow \delta = \frac{2.4 - 2.2}{1.1} = \frac{0.2}{1.1} \approx 0.1818

Now solve for $\rho_{X_1X_2}$:

\delta = \frac{\rho_{X_1X_2} \cdot 4 \cdot 7}{16} = \frac{28 \rho_{X_1X_2}}{16} = 1.75 \rho_{X_1X_2}

So: $`$0.18`18 = 1.75 \rho_{X_1X_2} \Rightarrow \rho_{X_1X_2} = \frac{0.1818}{1.75} \approx 0.1039

Thus, the correct answer is D. 0.1039.