Answer
$v=\sqrt{gy}$
Work Step by Step
First of all, we need to draw the force diagram of the ball as seen below.
Now we can apply Newton's second law on the ball in the $x$-direction which is the radial direction here.
$$\sum F_x=F_n\cos\theta=ma_r=m\dfrac{v^2}{R}$$
$$F_n\cos\theta= m\dfrac{v^2}{R}\tag 2$$
where $R$, from the two right triangles where one of them is part of the bigger one, is given by
$$\dfrac{R}{a}=\dfrac{y}{h}$$
Thus,
$$R=\dfrac{ya}{h}$$
Plugging into (2);
$$F_n\cos\theta= m\dfrac{v^2h}{ya}\tag 3$$
Now we need to find the normal force which is given by
$$\sum F_y=F_n\sin\theta-mg=ma_y=m(0)=0$$
Thus,
$$F_n\sin\theta=mg$$
$$F_n=\dfrac{mg}{\sin\theta}$$
Plugging into (3);
$$\dfrac{\color{red}{\bf\not} mg}{\sin\theta}\cos\theta= \color{red}{\bf\not} m\dfrac{v^2h}{ya} $$
$$ g\tan\theta= \dfrac{v^2h}{ya} $$
where $\tan\theta=\dfrac{h}{a}$;
$$ g\dfrac{\color{red}{\bf\not} h}{\color{red}{\bf\not} a}= \dfrac{v^2\color{red}{\bf\not} h}{y\color{red}{\bf\not} a} $$
Therefore,
$$\boxed{v=\sqrt{gy}}$$