Consider an ice cube of edge 1.0 cm placed in a gravity-free hall....

Volume of the ice cube is $V = a^3 = (1.0)^3 = 1.0 \text{ cm}^3$.

Since the difference in densities is ignored, the volume of water formed is also $V = 1.0 \text{ cm}^3$.

In a gravity-free space, the liquid takes a spherical shape to minimize its surface area due to surface tension.

Let the radius of the spherical water drop be $R$. Equating the volumes:

$\dfrac{4}{3}\pi R^3 = 1$

$R = \left(\dfrac{3}{4\pi}\right)^{1/3}$

The surface area of the spherical drop is given by $A = 4\pi R^2$.

$A = 4\pi \left(\dfrac{3}{4\pi}\right)^{2/3}$

$A = (4\pi)^{1 - 2/3} (3^2)^{1/3}$

$A = (4\pi)^{1/3} (9)^{1/3} = (36\pi)^{1/3} \text{ cm}^2$