Solve $f^{\prime}(x)=3 x^{2}-12 x+9=3(x-1)(x-3)$, then $f(x)$ is monotonically increasing on $(-\infty, 1)$, and monotonically decreasing on $(1,3)$, so $f(x)_{\text {max }}=f(1)=4$ and $f(4)=4$. Therefore, $\left\{\begin{array}{l}3-a^{2}<1<a, \\ a \leqslant 4\end{array} \Rightarrow \sqrt{2}<a \leqslant 4\right.$. Thus, the range of real number $a$ is $(\sqrt{2}, 4]$.
10. If the function $f(x)=x^{3}-6 x^{2}+9 x$ has a maximum value in $\left(3-a^{2}, a\right)$, then the range of real number $a$ is $\qquad$