Train motion in three phases: acceleration, constant velocity, deceleration Steps:

• Acceleration time: $t_\text{acc} = \frac{20}{0.20} = 100$ s; distance: $s_\text{acc} = \frac{1}{2} \times 0.20 \times 100^2 = 1000$ m.
• Deceleration time: $t_\text{dec} = \frac{20}{0.20} = 100$ s; distance: $s_{}$ m.
• Constant speed distance: $3000 - 1000 - 1000 = 1000$ m; time: $t_\text{const} = \frac{1000}{20} = 50$ s.
• Total time: $100 + 50 + 100 = 250$ s (note: closest option is C, but exact match absent; assumes approximation or problem intent).

Not enough information to match options exactly with given data.

Final answer: Not among choices