The aircraft takes 10s to go from one point (say A) to the other (say B) and AB subtends an angle of 30o at the point of observation (say O) .So, ABO forms a triangle.Now let OC be perpendicular bisector drawn from O to AB(i.e. actually the height).
Then in right trianle, OAC
AC = OC tan 15o = 3600 x 0.2679 = 964.44 m
so, AB = 2 x 964.44 = 1928.88 m
Now, the speed of aircraft , v = AB / t = 1928.88 / 10 = 192.888 m/s.