A solid glass sphere (n = 1.50) of radius 6.0 cm has a tiny air bubble inside…

Step 1. From the near side. $u = -3.0\ \text{cm}$. $\dfrac{n_2}{v} = \dfrac{n_2 - n_1}{R} + \dfrac{n_1}{u} = \dfrac{-0.50}{-6.0\ \text{cm}} + \dfrac{1.50}{-3.0\ \text{cm}} = 0.0833\ \text{cm}^{-1} - 0.500\ \text{cm}^{-1} = -0.417\ \text{cm}^{-1}$. So $v = -2.4\ \text{cm}$.

Step 2. The magnification. $m = \dfrac{1.50 \times (-2.4\ \text{cm})}{1.00 \times (-3.0\ \text{cm})} = 1.2$.

Step 3. From the far side. $u = -9.0\ \text{cm}$, and $R$ is still $-6.0\ \text{cm}$. $\dfrac{1.00}{v} = 0.0833\ \text{cm}^{-1} + \dfrac{1.50}{-9.0\ \text{cm}} = 0.0833\ \text{cm}^{-1} - 0.167\ \text{cm}^{-1} = -0.0833\ \text{cm}^{-1}$. So $v = -12\ \text{cm}$.

Step 4. The magnification. $m = \dfrac{1.50 \times (-12\ \text{cm})}{1.00 \times (-9.0\ \text{cm})} = 2.0$.