Weight of the bowl$=\mathrm{mg}=\text{V}\mathrm{\rho }\text{g}=\frac{2}{3}\mathrm{\pi }\left[\right(\frac{\text{D}}{2}{)}^3-\left(\frac{\text{d}}{2}{)}^3\right]\mathrm{\rho }\text{g}$, where $\mathrm{D}$ is the outer diameter, $\mathrm{d}$ is the inner diameter and $\mathrm{\rho }$ is the density of bowl.
Weight of the liquid displaced by the bowl${\text{=V}}^{\text{'}}\mathrm{\sigma }\text{g}=\frac{2}{3}\mathrm{\pi }{\left(\frac{\text{D}}{2}\right)}^3\mathrm{\sigma }\text{g}$, where $\mathrm{\sigma }$ is the density of the liquid.
For the flotation,

$\frac{2}{3}\pi (\frac{\text{D}}{\text{2}}{)}^3\mathrm{\sigma }\text{g}=\frac{2}{3}\mathrm{\pi }[(\frac{\text{D}}{2}{)}^3-(\frac{\text{d}}{2}{)}^3]\rho \text{g}$

By solving we get $\mathrm{d}=\text{0.98} \mathrm{m}$