Suppose the side length is $s$ cm, then the area is $s^2$ cm$^2$.
The radius of the circle is 1 cm, so the distance from the centre of the circle ($O$) to the vertices of the square that touch the circumference ($Q$ and $R$) is 1 cm.
This, the side length $s$ and the half-side length $\frac{1}{2}s$ are shown on the diagram:
Angle $OSR$ is a right angle (as it is in a square), so applying Pythagoras' theorem to triangle $OSR$ gives
$$\begin{align} s^2+(\frac{1}{2}s)^2 &= 1^2 \\
\Rightarrow s^2+\frac{1}{4}s^2 &= 1\\
\Rightarrow \frac{5}{4}s^2 &= 1\\
\Rightarrow s^2&=\frac{4}{5}\end{align}$$So the area of the square is $\dfrac{4}{5}$cm$^2$.