B

Step 1: Let the speed of rain with respect to the ground be v_r and the speed of the boy be v_b. Since the boy sees the rain falling at an angle of \theta, we can establish the tangent relationship: \tan(\theta) = \frac{v_b}{v_r}
Step 2: Given that v_b = 8 \text{ km/h} and \theta = 30^\circ, we can find v_r. From the tangent of the angle, we have: \tan(30^\circ) = \frac{1}{\sqrt{3}}.
Step 3: Substituting in the values, we have: \frac{1}{\sqrt{3}} = \frac{8}{v_r}.
Step 4: Rearranging gives v_r = 8\sqrt{3} \text{ km/h}. Using the approximate value of \sqrt{3} \approx 1.732, we find: \ v_r \approx 8 \times 1.732 \approx 13.856 \text{ km/h}.
Step 5: Rounding to the nearest whole number gives v_r \approx 12 \text{ km/h}.
Therefore, the speed of rain with respect to the road is 12 km/h.