A ball lens is a glass sphere of refractive index 1.60 and radius 1.5 mm…

Step 1. The first surface. The object is at infinity, so $\dfrac{n}{v_1} = \dfrac{n - 1}{R}$. We want $v_1$, so $v_1 = \dfrac{nR}{n - 1} = \dfrac{1.60 \times 1.5\ \text{mm}}{0.60} = 4.0\ \text{mm}$.

Step 2. The object for the second surface. The back surface is $2R = 3.0\ \text{mm}$ from the front. The rays are heading for a point $4.0\ \text{mm} - 3.0\ \text{mm} = 1.0\ \text{mm}$ beyond it. This is a virtual object, so $u_2 = +1.0\ \text{mm}$.

Step 3. The second surface. Light goes from glass to air: $n_1 = 1.60$, $n_2 = 1.00$ and $R_2 = -1.5\ \text{mm}$. $\dfrac{1.00}{v_2} = \dfrac{1.00 - 1.60}{-1.5\ \text{mm}} + \dfrac{1.60}{1.0\ \text{mm}} = 0.40\ \text{mm}^{-1} + 1.60\ \text{mm}^{-1} = 2.0\ \text{mm}^{-1}$. So $v_2 = 0.5\ \text{mm}$.

Step 4. From the centre. The focus is $0.5\ \text{mm}$ behind the back surface, so it is $1.5\ \text{mm} + 0.5\ \text{mm} = 2.0\ \text{mm}$ from the centre.