Acceleration of train , a = 20cm/s² = 0.2 m/s²
DECELERATION , d = - 100cm/s² =- 1 m/s²

Let the train start with acceleration for xkm or, 1000X m distances and deceleration for (2700 - 1000x) m [ I just change (2.7 - x) km into metre.]

so, equation of distance travelled by train in 1ST CASE :- when we assume train start from rest then, intial velocity, u = 0
use formula, v² = u² + 2aS
v² = 0 + 2 × 0.2 × 1000x = 400x ----(1)

equation of distance travelled by train in 2nd case :- intial velocity in this case = final velocity of 1st case = v.finally train will be rest .
so, final velocity in this case = 0
deceleration = d and distance is S' = (2700-1000x) m
so, 0 = v² + 2(-1) × (2700-1000x)
v² = 5400 - 2000x -----(2)

solve equations (1) and (2),
5400 - 2000x = 400x
=> 5400 = 2400x
=> x = 54/24 = 9/4 km

so, v² = 400 × 9/4 = 900
v = 30 m/s
now, use formula v = u + at
30 = 0 + 0.2t
t = 150 sec

again, time TAKEN in 2nd case
0 = 30 + (-1)t => t = 30 sec

hence, total time taken = 180 sec