1. **Problem statement:** A hollow sphere of mass $m_1 = m$ and radius $R$ rests on a frictionless horizontal surface. Inside it, a solid sphere of mass $m_2 = \frac{32}{7}m$ and radius $\frac{R}{2}$ is placed off-center. When the inner sphere oscillates to the lowest point inside the hollow sphere, we want to find the horizontal displacement of the hollow sphere relative to $R$. 2. **Key idea:** Since the system is isolated horizontally and the surface is frictionless, the center of mass (COM) of the system does not move horizontally. 3. **Define variables:** Let $x_1$ be the horizontal displacement of the hollow sphere's center, and $x_2$ be the horizontal displacement of the solid sphere's center relative to the hollow sphere. 4. **Center of mass condition:** The total horizontal COM position $X_{COM}$ is constant: $$ X_{COM} = \frac{m_1 x_1 + m_2 (x_1 + x_2)}{m_1 + m_2} = \text{constant} $$ 5. **Initial condition:** Initially, both spheres are at rest and aligned, so $x_1 = 0$ and $x_2 = 0$, thus $X_{COM} = 0$. 6. **At lowest point of inner sphere oscillation:** The inner sphere moves horizontally inside the hollow sphere by $x_2 = R - \frac{R}{2} = \frac{R}{2}$ (since it moves from the center to the bottom-left, horizontal displacement is $\frac{R}{2}$ to the left). 7. **Apply COM conservation:** $$ 0 = \frac{m x_1 + \frac{32}{7} m (x_1 + \frac{R}{2})}{m + \frac{32}{7} m} = \frac{m x_1 + \frac{32}{7} m x_1 + \frac{32}{7} m \frac{R}{2}}{m (1 + \frac{32}{7})} $$ 8. **Simplify numerator:** $$ m x_1 + \frac{32}{7} m x_1 + \frac{32}{7} m \frac{R}{2} = m x_1 \left(1 + \frac{32}{7}\right) + \frac{32}{7} m \frac{R}{2} $$ 9. **Cancel denominator and solve for $x_1$:** $$ 0 = x_1 \left(1 + \frac{32}{7}\right) + \frac{32}{7} \cdot \frac{R}{2} $$ $$ x_1 = - \frac{\frac{32}{7} \cdot \frac{R}{2}}{1 + \frac{32}{7}} = - \frac{\frac{32}{7} \cdot \frac{R}{2}}{\frac{39}{7}} = - \frac{32}{7} \cdot \frac{R}{2} \cdot \frac{7}{39} = - \frac{32 R}{2 \cdot 39} = - \frac{16 R}{39} $$ 10. **Interpretation:** The hollow sphere moves horizontally by $\boxed{-\frac{16}{39} R}$, i.e., to the left by $\frac{16}{39} R$ relative to its initial position. **Final answer:** $$ \text{Horizontal displacement of hollow sphere} = - \frac{16}{39} R $$ This means the hollow sphere shifts left by $\frac{16}{39} R$ when the inner sphere is at the lowest point.