The wall is ¼ of the range away. 
The maximum height is ½ of the range away. 
So it takes as long to get from the throw point to the wall as it takes to get from the wall to maximum height. 
If the total flight time is T, then the vertical velocity Vv is 
Vv at t = 0, 
½*Vv at t = T/4, 
0 at t = T/2, 
-½Vv at t = 3T/4, and 
-Vv at t = T. 
Then the average velocity during the first interval is 3Vv/4, and the average velocity during the second interval is Vv/4. 
Then the vertical distance traveled during the first interval is 3 times the vertical distance during the second interval. 
Since the vertical distance traveled during the first interval was 3 m, then the vertical distance traveled during the second interval was 1 m, and the maximum height is 4 m. 

range = (V²sin(2Θ))/g → 24m = V²sin(2Θ) / g 
max height = (V·sinΘ)² / (2g) → 4m = V²sin²Θ / 2g 
Divide range by max height: 
24m / 4m = 6 = 2sin(2Θ) / sin²Θ 
Use trig identity sin(2Θ) = 2sinΘcosΘ: 
6 = 4sinΘcosΘ / sin²Θ = 4cosΘ / sinΘ = 4 / tanΘ 
tanΘ = 4/6 = 2/3 
Θ = arctan(2/3) = 33.7º 
assuming the ball just clears the wall. 
 
						

							
Use the equation of trajectory for solvingvthe question:-
Taking y as 3 and x as 6 and then find the value of theta
Use the formula y=x tan theta(1-X/R)...
						

							
Use eqn of trajectory y = x(1-x/R)tan# taking x = 6 and y = 3 and r as 24 and find the value of # it would come out as arctan(2/3) which is the required angle of projection.