(a) Find the average value of the function $f(x)=3\sqrt{x}$ on the interval $[0,4]$.
$$f_{ave}=$$
(b) Find $c$ such that $f_{\text{ave }}=f(c)$.
$$c=$$

Step 1 ：The average value of a function $f(x)$ on the interval $[a,b]$ is given by the formula: $$f_{ave}=\frac{1}{b-a}{\int }_a^{b}f(x)dx$$

Step 2 ：In this case, $f(x)=3\sqrt{x}$, $a=0$ and $b=4$. So we need to calculate the integral of $f(x)$ from $0$ to $4$ and then divide it by $4-0=4$.

Step 3 ：The average value of the function $f(x)=3\sqrt{x}$ on the interval $[0,4]$ is 4.

Step 4 ：Now, we need to find the value of $c$ such that $f_{ave}=f(c)$. This means we need to solve the equation $f(c)=f_{ave}$ for $c$.

Step 5 ：The value of $c$ such that $f_{ave}=f(c)$ is $\frac{16}{9}$.

Step 6 ：Final Answer: The average value of the function $f(x)=3\sqrt{x}$ on the interval $[0,4]$ is $\boxed{{\displaystyle 4}}$ and the value of $c$ such that $f_{ave}=f(c)$ is $\boxed{{\displaystyle \frac{16}{9}}}$.