How to find acceleration only having speed and distance?

This question can be answered only if you assume that the aircraft started at a standstill, and accelerated at a constant rate.

You can find the time with this formula:
t = (2d) / v
...which only works when the object starts at a standstill and accelerates at a constant rate. Once you have the time, just divide the velocity by it, and you've got your answer.

I don't think U can find the "minimum acceleration" since IF the plane doesn't accelerate uniformly there is no way of finding the minimum acceleration with the data given.

However, if U had stated "average acceleration" THAT can be determined by definition regardless of the way the plane actually accelerated down the runway.

initial speed = Vo = 0
final speed = Vf = 125 km/h
Vavg = average speed of take off = 125/2 = 62.5 km/h = 17.36 m/s
take off distance = 277 m
T = time to take off = 277/Vavg = 277/17.36 = 15.96 = 16.0 s
size of average acceleration = (Vf - Vo)/T = Vf/16 = __________  U can solve* this :>)
*don't forget to convert Vf to BASIC units of speed in the SI system before calculating.

125km/h = 125 x 1000/3600 = 34.72m/s

Here are 2 methods.
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Since it starts from rest, average speed during take-off = (0+34.72)/2 = 17.36m/s

Time = distance/average speed = 277/17.36 = 15.96s

Acceleration = change in speed/time = 34.72/15.96 = 2.18m/s²
_____________________

The other method is to use the standard formula v² = u² +2as (or vf² = vi² + 2as if you use those symbols)

v² = u² +2as
34.72² = 0² + 2a x 277
a = 1205/(2x277)
= 2.18m/s²

EDIT. The points made in the other answers are correct.  The question should state that the plane starts from rest and that you are to find the minimum average acceleration (or that acceleration is constant).

(Vf)² = (Vi)² + 2?a?d
where
Vf = final velocity = 125 km/h = 34.72 m/s
Vi = initial velocity = 0 m/s
a = acceleration = ?
d = distance = 277 m

so

34.72² = 0² + 2?a?277
1205.6 = 554a
a = 2.2 m/s²