Answer
See the detailed answer below.
Work Step by Step
Since the water moves as one unit, so we can choose an arbitrary molecule on the surface of the water and work on it, as seen below.
Now we need to apply Newton's second law in both directions.
$$\sum F_r=F_n\cos\theta=ma_r=m\dfrac{v^2}{r}$$
where $v=\omega r$; Thus,
$$F_n\cos\theta =m\dfrac{\omega^2r^2}{r}=m\omega^2r$$
$$F_n\cos\theta=m\omega^2r\tag 1$$
$$\sum F_z=F_n\sin\theta-mg=ma_z=m(0)=0$$
thus,
$$F_n=\dfrac{mg}{\sin\theta}$$
Plugging into (1);
$$\dfrac{mg}{\sin\theta}\cos\theta=m\omega^2r $$
$$ g\cot\theta= \omega^2r $$
$$ g = \dfrac{\omega^2r}{\cot\theta} =\omega^2r\tan\theta$$
$$ g = \omega^2r\tan\theta\tag 2$$
We know the parabola curve is given by $z=ar^2$ and hence,
$$\dfrac{dz}{dr}=2ar $$
And at point A, $dz/dr$ is the slope of the line that is tangential to the parabola curve at this point A.
$$\dfrac{dz}{dr}=2ar={\rm Slope_A}$$
Noting that the slope of this line is perpendicular to the normal force and this slope is given by
$$\dfrac{dz}{dr}=2ar={\rm Slope_A}=\tan\phi$$
where $\phi=90^\circ-\theta$
$$ 2ar =\tan(90^\circ-\theta)=\dfrac{1}{\tan\theta}$$
Solving (2) for $\tan\theta$ and plug it here;
$$ 2a\color{red}{\bf\not} r = \dfrac{\omega^2 \color{red}{\bf\not} r}{g}$$
Thus,
$$a=\dfrac{\omega^2 }{2g}$$
Plug it into the parabola formula of $z=ar^2$;
$$\boxed{z=\left(\dfrac{\omega^2 }{2g}\right)r^2}$$
