1. **State the problem:** We have a circuit with a voltage source $E=10.5$ V and resistors $R_1=860\ \Omega$, $R_2=3.68\ \text{k}\Omega=3680\ \Omega$, $R_3=854\ \Omega$, and $R_4=3.85\ \text{k}\Omega=3850\ \Omega$. We need to find the voltage $V_x$ at the node between $R_2$ and $R_3$ using the voltage divider rule. 2. **Analyze the circuit:** The circuit has a series path: $E$ connected to $R_1$, then branching to $R_2$ upwards to $V_x$, then down through $R_3$ and $R_4$ back to the negative terminal of $E$. 3. **Simplify the circuit:** Resistors $R_3$ and $R_4$ are in series, so their equivalent resistance is: $$R_{34} = R_3 + R_4 = 854 + 3850 = 4704\ \Omega$$ 4. **Find total resistance in the main loop:** The total resistance is $R_1$ in series with the parallel combination of $R_2$ and $R_{34}$: $$R_{p} = \frac{R_2 \times R_{34}}{R_2 + R_{34}} = \frac{3680 \times 4704}{3680 + 4704} = \frac{17310720}{8384} \approx 2064.5\ \Omega$$ Total resistance: $$R_{total} = R_1 + R_p = 860 + 2064.5 = 2924.5\ \Omega$$ 5. **Calculate total current $I$ in the circuit:** $$I = \frac{E}{R_{total}} = \frac{10.5}{2924.5} \approx 0.00359\ \text{A} = 3.59\ \text{mA}$$ 6. **Calculate voltage across the parallel branch $V_p$:** $$V_p = I \times R_p = 0.00359 \times 2064.5 \approx 7.41\ \text{V}$$ 7. **Calculate voltage $V_x$ across $R_2$ using voltage divider rule:** Since $R_2$ and $R_{34}$ are in parallel, voltage across $R_2$ is equal to $V_p$: $$V_x = V_p = 7.41\ \text{V}$$ 8. **Express $V_x$ in engineering notation rounded to three significant digits:** $$V_x = 7.41\ \text{V}$$ **Final answer:** $$\boxed{V_x = 7.41\ \text{V}}$$