Answer
$0$
Work Step by Step
$y = \sin^{-1}{x} \hspace{15pt} \to \hspace{15pt} x = \sin{y}$
$\text{where } \hspace{15pt} -1 \leq x \leq 1 \hspace{15pt} \text{and} \hspace{15pt} -\dfrac{\pi}{2} \leq y \leq \dfrac{\pi}{2}$
Let $\theta = \sin^{-1} {0} \hspace{20pt} -\dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2}$
Since $\theta = \sin^{-1} {0}$ ,then
$\sin{\theta} = 0$
Referring to Table 1 on page 466:
The only angle in the interval$ \left[-\dfrac{\pi}{2},\dfrac{\pi}{2} \right] $whose sine is $0$ is $0$
Therefore,
$\sin^{-1} {0} = 0$