Answer
$3$ cm below the surface.
Work Step by Step
In this situation, the object, which is the air bubble, is behind the refracted surface of the sphere and it faces a concave surface.
Thus, $R=-3$ cm.
And since the air bubble is just in the middle, $s=3$ cm.
Recalling for spherical surfaces that object and image distances are related by
$$\dfrac{n_1}{s}+\dfrac{n_2}{s'}=\dfrac{n_2-n_1}{R}$$
where $n_1$ here is for zircon and it is 2.18 [see table 23.1], while $n_2$ is for air and is it 1.0.
Now we need to plug the known and solve for $s'$,
$$\dfrac{2.18}{3}+\dfrac{1}{s'}=\dfrac{1-2.18}{-3}$$
Hence,
$$s'=\color{red}{\bf-3}\;{\rm cm}=R$$
This means that when the bubble is at the center, it appears to be exactly at the center.
