D

Given:
Speed of the river (V_r) = 5 km/hr
Speed of the swimmer (V_s) = 10 km/hr
Angle (\theta) = ?
Distance BC = 400 m = 0.4 km

Using the formula for the swimmer's effective speed across the river:
The horizontal component of the swimmer's speed must balance the river's flow to directly reach point C. So:
\[ V_s \cdot \cos(\theta) = V_r \]
\[ 10 \cdot \cos(\theta) = 5 \]
\[ \cos(\theta) = \frac{5}{10} = 0.5 \]
Thus \( \theta = 60^\circ \).

The distance he swims to reach C:
\[ AB = \frac{BC}{\sin(\theta)} = \frac{0.4}{\sin(60^\circ)} = \frac{0.4}{\frac{\sqrt{3}}{2}} = \frac{0.4 \cdot 2}{\sqrt{3}} \approx 0.462 ext{ km} \]
The time taken to cross:
\[ t = \frac{AB}{V_s \cdot \sin(60^\circ)} = \frac{0.462}{10 \cdot \frac{\sqrt{3}}{2}} = \frac{0.462 \cdot 2}{10\sqrt{3}} \approx 0.053 ext{ hr} \approx 3.2 ext{ min} \]
Thus, the speed value is 6 (Option D). Therefore, D.