The total distance covered is given by the area under the speed curve and can be calculated by integrating the speed curve equation in the given range. In the question, exact speed function is not given, but the speed at different time instants is given.
Question is asking the distance covered in 20 minutes. According to the question, the motion starts at t = 0 and ends at t = 20. So the speed function has to be integrated in the range t=0 to t=20. Therefore an additional point (t=0) has to be considered while integrating. After adding t=0, number of points will be 11 and number of intervals will be 10. And the answer will be 309.33.
Distance
$S = \frac{(b-a)}{3n}\left [ f(0) + 4 \left( f(2) + f(6) + f(10) + f(14) + f(18) \right) \\+ 2\left ( f(4) + f(8) + f(12) + f(16) + f(20) \right)\right]$
$= \frac{2}{3} \left[0 + 4\left(10 + 25 + 32 + 11 + 2\right) \\+ 2\left(18 + 29 + 20 + 5 \right)\right]$
$ = 309.33$
Ref: http://www.saylor.org/site/wp-content/uploads/2011/11/ME205-7.2-TEXT2.pdf