1. **State the problem:** Find the mass of block $B$ in a pulley system in static equilibrium, given the mass of block $D$ and the geometry of the system. 2. **Tension in the rope:** Since the rope is continuous and pulleys are ideal, tension $T$ is constant. At block $D$, equilibrium gives: $$ T = m_D g = 5 g $$ 3. **Geometry and angle $\theta$:** Total rope length $L = 6$ m, vertical segment $a = 2.6$ m, horizontal distance $b = 0.333$ m. Length of segments $AB$ and $CB$: $$ L_R = \frac{L - a}{2} = \frac{6 - 2.6}{2} = 1.7 \text{ m} $$ Calculate $\cos \theta$: $$ \cos \theta = \frac{b/2}{L_R} = \frac{0.1665}{1.7} \approx 0.097941 $$ 4. **Calculate $\sin \theta$ using Pythagorean identity:** $$ \sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - (0.097941)^2} = \sqrt{0.9904076} \approx 0.995192 $$ 5. **Equilibrium at pulley $B$:** Vertical forces balance: $$ 2 T \sin \theta = m_B g $$ Substitute $T = m_D g$ and cancel $g$: $$ m_B = 2 m_D \sin \theta = 2 \times 5 \times 0.995192 = 9.95192 $$ 6. **Final answer rounded to three decimals:** $$ m_B = 9.952 \text{ kg} $$