Ans is A.
Exact calculations may make this one lengthy so the approach you should follow is one of intuitive thinking i.e. in case of a sqaure Moment of Inertia ~ (L)^2 and torque of friction ~ (L) ( conside a square having side 2L then for every element situated at 'r' distance from centre in square of side L there would be a corresponding element situated at '2r' in it with same force acting on it.. Hence torque would be dobuled)
So alpha ~ 1/L and t ~ L and hence t1/t2=L/2L=1/2...
But you would surely think where in the heck is the Square with side length 2L mentioned in the question and for that let me give you a simple hint a plate of side 2L spinning around the centre could be considered as 4 plates of side L spinning around their corners and each of these small plates ( of side L) will experience equal torque(T/4 if that experienced by complete square is T)and have equal moment of inertia(I/4 if that of complete square is I). Hence the angular acceleration of plate of side L spinning around it's vertex would be same as that of a plate of side 2L spinning around the centre.