The radius of Sm³⁺ is 95.8 pm and that of Yb³⁺ is 86.8 pm…

Step 1. Sm. $95.8\ \text{pm} = 0.0958\ \text{nm}$, so $\dfrac{3}{0.0958\ \text{nm}} = 31.3\ \text{charges/nm}$.

Step 2. Yb. $86.8\ \text{pm} = 0.0868\ \text{nm}$, so $\dfrac{3}{0.0868\ \text{nm}} = 34.6\ \text{charges/nm}$.

Step 3. The rise. $\dfrac{34.6 - 31.3}{31.3} \times 100 = 10.4\ \%$ (using the unrounded values, $34.56$ and $31.32$).

Step 4. The base. The higher ionic potential belongs to $\text{Yb}^{3+}$. It pulls the electrons of the oxygen harder, the Yb-O bond is more covalent, and $\text{OH}^-$ is given up less readily. So $\text{Yb(OH)}_3$ is the weaker base.