Answer
a) $19.6 x$
b) $3.92\;\rm m/s^2$
Work Step by Step
a)
Let's take the slope of the ball at the place where it lies on the given figure.
The forces exerted on the ball are the normal force at an angle $\theta$ that is equal to the angle of the slope with horizontal $\theta$, as you see below.
We know that the slope is given by
$${\rm Slope}=\dfrac{dy}{dx}=\tan\theta\tag 1$$
And we know that the path equation is given by
$$y=x^2$$
Hence,
$$\dfrac{dy}{dx}=2x$$
Plugging into (1);
$$\tan\theta=2x\tag 2$$
The net force exerted on the ball in the $y $-direction is zero
$$\sum F_{y }=F_n \cos\theta-mg=0$$
$$F_n=\dfrac{mg}{ \cos\theta}\tag 3$$
The net force exerted on the ball in the $x $-direction is
$$\sum F_{x }=F_n\sin\theta=ma_x$$
Hence,
$$F_n\sin\theta=ma_x$$
Plugging from (3);
$$ \dfrac{mg}{ \cos\theta}\; \sin\theta =ma_x$$
Thus,
$$a_x=g\tan\theta$$
Plugging from (2);
$$\boxed{a_x=2g x}$$
$$\boxed{a_x={\bf19.6} x}$$
b)
$$a_x=19.6\cdot 20\times 10^{-2}=\color{red}{\bf 3.92}\;\rm m/s^2$$
