smc

$\sqrt{{\pi }^2}=\pi$ is not rational, so eliminate choices $(B)$, $(C)$, and $(E)$. Note that $(-1{)}^3=(-1)$, so $\sqrt[3]{-1}=-1$ is rational. We have found one rational number, so we can eliminate choice $(A)$. The answer is $\boxed{\text{only the fourth}}$.