Answer
$4.78\;\rm m/s$
Work Step by Step
First of all, we need to find the first bouncing angle.
We can see, from the figure below, that the falling angle (relative to the tilted surface and to the left from the vertical line) is 70$^\circ$.
Hence, the first bounce angle must be 70$^\circ$ as well relative to the tilted surface, but we need its angle relative to the horizontal line which is then 50$^\circ$. For more clarification see the figure below.
Now we need to find $v_0$.
We chose the first bounce point to be our origin at which $$(x_0,y_0)=(0,0)\tag 1$$
And hence, the second bounce point is given by
$$(x_1,y_1)=(3,-h)$$
$$(x_1,y_1)=(3,-3\tan20^\circ)\tag 2$$
and $h$ from the figure below is given by
$$\tan20^\circ =\dfrac{h}{3}$$
Thus,
$$h=3\tan20^\circ$$
We know that $v_x$ is constant, so
$$v_x=v_0\cos50^\circ=\dfrac{x_1-x_0}{t}$$
Thus,
$$t=\dfrac{3}{v_0\cos50^\circ}\tag 3$$
We know that $v_y$ changes since the ball moves under the free-fall acceleration, so $a_y=-g$
$$y_1=y_0+v_yt-\frac{1}{2}gt^2$$
$$-3\tan 20^\circ=0+v_0\sin50^\circ t-4.9t^2$$
Plugging from (3);
$$-3\tan 20^\circ= v_0\sin50^\circ \dfrac{3}{v_0\cos50^\circ}-4.9\left[\dfrac{3}{v_0\cos50^\circ}\right]^2$$
$$-3\tan 20^\circ=3\tan50^\circ - \dfrac{44.1}{v_0^2\cos^250^\circ} $$
$$\dfrac{44.1}{v_0^2\cos^250^\circ}=3\tan50^\circ+3\tan 20^\circ $$
$$\dfrac{44.1}{\cos^250^\circ(3\tan50^\circ+3\tan 20^\circ )}=v_0^2 $$
$$v_0=\sqrt{\dfrac{44.1}{\cos^250^\circ(3\tan50^\circ+3\tan 20^\circ )}}$$
Thus,
$$v_0=\color{red}{\bf 4.78}\;\rm m/s$$
