Answer
$$\mathcal{L}^{-1}\left\{F(s)\right\}=2t^2e^t$$
Work Step by Step
We see that $\dfrac{4}{(s-3)^3}$ most closely resembles the form of of #11 in Table 6.2.1. Let $n=2$ for that entry and let $a=1$. Then, we need to do some algebra so that the numerator is equivalent to $n!$, or in this case the numerator is equal to $2!=2$.
\begin{align*}
\mathcal{L}^{-1}\left\{\dfrac{4}{(s-1)^3}\right\}&=2\cdot \mathcal{L}^{-1}\left\{\dfrac{2}{(s-1)^{2+1}}\right\}\\
\\
&=2\cdot \mathcal{L}^{-1}\left\{\dfrac{(2)!}{(s-1)^{2+1}}\right\}\\
\\
&=2\cdot t^{2}e^{1t}\\
\\
&=\boxed{\color{blue}{2t^2e^t}}\\
\end{align*}
