The largest interval in which the function $f\left(x\right)=3\sin \left(\sqrt{\frac{{\pi }^2}{16}-x^2}\right)$  assumes values is :

f  is defined only if $\frac{{\pi }^2}{16}-x^2\ge 0$

i.e. $x^2\le \frac{{\pi }^2}{16}\Rightarrow x\le \frac{\pi }{4}$  or $x\ge \frac{-\pi }{4}$

$\Rightarrow \frac{-\pi }{4}\le x\le \pi /4$ .

Since $f\left(-\pi /4\right)=0,f\left(\pi /4\right)=0$  and $f\left(0\right)=3\sin \frac{\pi }{4}=\frac{\sqrt{3}}{2}$ .

$\therefore$  f takes values in the interval $\left[0,\frac{3}{\sqrt{2}}\right]$ .