A body is floating in a liquid partly immersed. If the body and the liquid are taken to the Moon, the body will:

For a body floating in a liquid, the fraction of the volume submerged ($V_{\text{in}}$) out of total volume ($V$) is determined by the equilibrium between buoyant force and weight:
$$V_{\text{in}}\cdot {\rho }_{\text{liquid}}\cdot g=V\cdot {\rho }_{\text{body}}\cdot g$$

Dividing both sides by $g$:

$$\frac{V_{\text{in}}}{V}=\frac{{\rho }_{\text{body}}}{{\rho }_{\text{liquid}}}$$

Since the acceleration due to gravity ($g$) cancels out on both sides, the fraction of the body submerged depends entirely on the densities of the body and the liquid, which remain constant. Therefore, even though gravity on the Moon is ${\frac{1}{6}}^{\text{th}}$ of that on Earth, the body will continue to float exactly as before.