Weight of the bowl = mg

= $V \rho g = \frac{4}{3} \pi \left[ \left(\frac{D}{2}\right)^3 - \left(\frac{d}{2}\right)^3 \right] \rho g$

where D = Outer diameter,

d = Inner diameter

\rho = Density of bowl

Weight of the liquid displaced by the bowl

$= V\sigma g = \frac{4}{3} \pi \left(\frac{D}{2}\right)^3 \sigma g$

where $\sigma$ is the density of the liquid.

For the flotation $\frac{4}{3}\pi\left(\frac{D}{2}\right)^3 \sigma g = \frac{4}{3}\pi \left[\left(\frac{D}{2}\right)^3 - \left(\frac{d}{2}\right)^3\right]\rho g$

⇒ ⇒ $\left(\frac{1}{2}\right)^3 \times 1.2 \times 10^3 = \left[\left(\frac{1}{2}\right)^3 - \left(\frac{d}{2}\right)^3\right] 2 \times 10^4$

By solving we get d = 0.98 m.