Given vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ satisfy $|\overrightarrow{a}|=3$, $|\overrightarrow{b}|=8$, and $|\frac{5}{3}\overrightarrow{a}-\overrightarrow{b}|=7$, we are tasked with finding $\overrightarrow{a} \cdot \overrightarrow{b}$.

First, we note the given conditions:
1. $|\overrightarrow{a}|=3$
2. $|\overrightarrow{b}|=8$
3. $|\frac{5}{3}\overrightarrow{a}-\overrightarrow{b}|=7$

We use the property that $|\overrightarrow{x}|^2 = \overrightarrow{x} \cdot \overrightarrow{x}$ to express the square of the magnitude of the vector $\frac{5}{3}\overrightarrow{a}-\overrightarrow{b}$ as follows:

$$\left|\frac{5}{3}\overrightarrow{a}-\overrightarrow{b}\right|^2 = \left(\frac{5}{3}\overrightarrow{a}-\overrightarrow{b}\right) \cdot \left(\frac{5}{3}\overrightarrow{a}-\overrightarrow{b}\right)$$

Expanding the dot product on the right-hand side gives:

$$\left|\frac{5}{3}\overrightarrow{a}-\overrightarrow{b}\right|^2 = \frac{25}{9}\overrightarrow{a} \cdot \overrightarrow{a} - \frac{10}{3}\overrightarrow{a} \cdot \overrightarrow{b} + \overrightarrow{b} \cdot \overrightarrow{b}$$

Substituting the given magnitudes of $\overrightarrow{a}$ and $\overrightarrow{b}$, and the magnitude of $\frac{5}{3}\overrightarrow{a}-\overrightarrow{b}$, we get:

$$49 = \frac{25}{9} \cdot 9 - \frac{10}{3}\overrightarrow{a} \cdot \overrightarrow{b} + 64$$

Simplifying this equation:

$$49 = 25 - \frac{10}{3}\overrightarrow{a} \cdot \overrightarrow{b} + 64$$

Bringing all terms involving $\overrightarrow{a} \cdot \overrightarrow{b}$ to one side and constants to the other gives:

$$- \frac{10}{3}\overrightarrow{a} \cdot \overrightarrow{b} = 49 - 25 - 64$$

Simplifying the right-hand side:

$$- \frac{10}{3}\overrightarrow{a} \cdot \overrightarrow{b} = -40$$

Solving for $\overrightarrow{a} \cdot \overrightarrow{b}$:

$$\overrightarrow{a} \cdot \overrightarrow{b} = \frac{-40}{-\frac{10}{3}} = 12$$

Therefore, the answer is $\boxed{C}$.