Answer
$ 34.3^\circ$
Work Step by Step
We know that the tilted surface is frictionless which means that the only force exerted on the ball is its weight components (due to Earth's gravitational pull).
We chose up, as you see below, the tilted surface to be our positive $y$-direction; see the first figure below that shows a side view.
And we chose right to be our positive $x$-direction; see the second figure below.
Now it is similar to a ball fired on air at an angle of $\theta$.
We know that the horizontal distance that should traveled by the ball is 2.5 m and we know that $v_x$ is constant.
So, the time of the trip is given by
$$v_x=\dfrac{\Delta x}{t}=v_0\cos\theta$$
$$t=\dfrac{2.5}{3\cos\theta}\tag 1$$
We also know that the net $y$-distance traved is zero since it starts and ends at the same level.
So,
$$\overbrace{y_f}^{0}=\overbrace{y_i}^{0}+v_yt-\frac{1}{2}a_yt^2$$
Thus,
$$v_yt=v_0\sin\theta t=\frac{1}{2}a_yt^2$$
From the figure below, $a_y=g\sin20^\circ $
$$ v_0\sin\theta =\frac{1}{2}g\sin20^\circ t $$
Thus,
$$\sin\theta=\dfrac{g\sin20^\circ t }{2 v_0}$$
Plugging from (1)
$$\sin\theta=\dfrac{g\sin20^\circ }{2 v_0}\left(\dfrac{2.5}{3\cos\theta}\right) $$
$$ \sin\theta\cos\theta=\dfrac{2.5g\sin20^\circ }{6v_0}$$
$$ 2\sin\theta\cos\theta=\dfrac{2\cdot2.5g\sin20^\circ }{6v_0}$$
$$ \sin2\theta =\dfrac{2\cdot2.5\cdot 9.8\sin20^\circ }{6\cdot 3}$$
$$ 2\theta =\sin^{-1}\left[\dfrac{2\cdot2.5\cdot 9.8\sin20^\circ }{6\cdot 3}\right]=68.6^\circ$$
Therefore,
$$\theta=\color{red}{\bf 34.3^\circ}$$