Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
This is not a thin lens problem, but we will treat each edge of the crystal ball as an independent lens where the image from the forward edge is the object of the backward edge.
Hence, the position of the image created by the forward edge is given by
$$\dfrac{n_1}{s_1}+\dfrac{n_2}{s_1'}=\dfrac{n_2-n_1}{R}$$
where the forward edge is convex toward the object, so $R=+5$ cm, $n_1=n_{air}=1$, and $n_2=n_{glass}=1.5$,
Solving for $s_1'$,
$$ s_1' =n_2\left[ \dfrac{n_2-n_1}{R}-\dfrac{n_1}{s_1}\right]^{-1}$$
$$ s_1' =1.5\left[ \dfrac{1.5-1}{5}-\dfrac{1}{6}\right]^{-1}$$
$$ s_1' =\bf-22.5\;\rm cm$$
This means that the image is on the same side of the object far from the forward side by a distance of 22.5 cm.
This virtual image is the object of the backward surface which is concave toward the object. And then its distance from this surface is
$s_2=22.5+10=32.5$ cm, and $R=-5$ cm since it is concave toward this object (the first virtual image).
Thus,
$$\dfrac{n_2}{s_2}+\dfrac{n_1}{s_2'}=\dfrac{n_1-n_2}{R}$$
Solving for $s_2'$,
$$ s_2' =n_1\left[\dfrac{n_1-n_2}{R}-\dfrac{n_2}{s_2}\right]^{-1}$$
$$ s_2' =(1)\left[\dfrac{1-1.5}{-5}-\dfrac{1.5}{32.5}\right]^{-1}$$
$$ s_2' =\bf18.6\;\rm cm$$
This means that the final image is at a distance of 18.6 cm from the backward side of the ball which means its distance from the center of the ball is given by
$$\Delta x=18.5+R=18.6+5=\color{red}{\bf 23.6}\;\rm cm$$
$$\color{blue}{\bf [b]}$$
See the figure below.
$$\color{blue}{\bf [c]}$$
In this case, when we place a thin lens at the center of the ball,
The position of the object is then $s_1 =6+5=11\;\rm cm$.
The position of the final image $s_1'=\Delta x=23.6\;\rm cm$,
Thus the focal length is then given by
$$\dfrac{1}{s_1}+\dfrac{1}{s_1'}=\dfrac{1}{f}$$
$$f=\left[ \dfrac{1}{s_1}+\dfrac{1}{s_1'} \right]^{-1} $$
$$f=\left[ \dfrac{1}{11}+\dfrac{1}{23.6} \right]^{-1} $$
$$f=\color{red}{\bf 7.5}\;\rm cm$$
