here train is moving in upward direction on an inclined plane having angle of inclination = 30 deg.

acceleration of the train = a = g/2

here component of acceleration in upward direction i.e. normal to the horizontal is (g/2)sin30 and along the horizontal it is (g/2)cos30

let string makes an angle  with the normal to the celing of train.

Tension in the string = T

Resolving the tension along two components as follows:

(a) Vertical to the ground

T cos( - 30) = m[g +(g sin30)/2]     (1)

(b) Along the ground that is in the horizontal direction

T sin( - 30) = mg (cos30)/2     (2)

Dividing eq. (2) by (1) we obtain,

tan( - 30) = [mg (cos30)]/m[g +(g sin30)/2] = 3 /5

or (tan - tan30)/(1+tantan30) = 3 /5

or tan = 2/3

or  = tan^-12/3 

Now to obtain tension in the string square and add then equate LHS and RHS of eq. (1) and (2)

by doing so we obtain and simplifing we obtain

T = 57