The total kinetic energy of the ball at the bottom of the curve can be calculated as follows:

$\begin{array}{rcl}K& =& \frac{1}{2}M{v}^2+\frac{1}{2}·\frac{2}{3}M{R}^2{\omega }^2\\ & =& \frac{1}{2}M{v}^2+\frac{1}{3}M{\left(R\omega \right)}^2\\ & =& \frac{1}{2}M{v}^2+\frac{1}{3}M{v}^2 \left[\because v=\omega R\right]\\ & =& \frac{5}{6}M{v}^2 ...\left(1\right)\end{array}$

The formula to calculate the potential energy acquired by the ball when it attains the maximum height is given by

$U=Mgh ...\left(2\right)$

Equate equation (1) and (2) and simplify to obtain the required height.

$\begin{array}{rcl}Mgh& =& \frac{5}{6}M{v}^2\\ & \Rightarrow & h=\frac{5v^2}{6g} ...\left(3\right)\end{array}$

Substitute the values of the known parameters into equation (3) to calculate the required height attained by the ball.
$\begin{array}{rcl}h& =& \frac{5\times {\left(3 \mathrm{m} {\mathrm{s}}^{-1}\right)}^2}{6\times 10 \mathrm{m} {\mathrm{s}}^{-2}}\\ & =& 0.75 \mathrm{m}\times \frac{100 \mathrm{cm}}{1 \mathrm{m}}\\ & =& 75 \mathrm{cm}\end{array}$