The angular momentum of a rotating body is ' $L$ '. When the frequency of rotating body is tripled and its kinetic energy is made one-third, the new angular momentum becomes

The angular momentum $L$ and rotational kinetic energy $K$ of a rotating body are related by its angular velocity $\omega = 2\pi f$, where $f$ is the frequency. Starting from the definition of rotational kinetic energy, $K = \frac{1}{2} I \omega^2$, and substituting the angular momentum relation $L = I \omega$ to eliminate the moment of inertia $I$, we obtain $I = \frac{L}{\omega}$.

Substituting into the kinetic energy expression gives $K = \frac{1}{2} \cdot \frac{L}{\omega} \cdot \omega^2 = \frac{1}{2} L \omega$. Replacing $\omega$ with $2\pi f$ yields $K = \frac{1}{2} L (2\pi f) = \pi L f$, which rearranges to $L = \frac{K}{\pi f}$.

For the initial state, $L_1 = \frac{K_1}{\pi f_1}$. When the frequency is tripled to $f_2 = 3f_1$ and the kinetic energy is reduced to one-third, $K_2 = \frac{1}{3} K_1$, the new angular momentum becomes $L_2 = \frac{K_2}{\pi f_2} = \frac{\frac{1}{3} K_1}{\pi (3f_1)} = \frac{1}{9} \cdot \frac{K_1}{\pi f_1} = \frac{1}{9} L_1$.

Final answer: $\boxed{\text{A}}$