The mass of the ball is\(m = 0.270\;{\rm{kg}}\).

The radius of the circle is\(r = 1.35\;{\rm{m}}\).

The angular speed is \(\omega = 10.4\;{\rm{rad/s}}\).

In this problem, use the product of the moment of inertia of the ball and its angular speed to evaluate the angular momentum.

The relation of angular momentum can be written as:

\(\begin{aligned}{l}L &= I\omega \\L &= \left( {m{r^2}} \right)\omega \end{aligned}\)

Here, Iis the moment of inertia of the ball.

On plugging the values in the above relation, you get:

\(\begin{aligned}{l}L &= \left( {0.270\;{\rm{kg}} \times {{\left( {1.35\;{\rm{m}}} \right)}^2}} \right)\left( {10.4\;{\rm{rad/s}}} \right)\\L &= 5.1\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\end{aligned}\)

Thus, \(L = 5.1\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\) is the required angular momentum.