\(AA'\perp A'B\),so AA' and the plane A'B'C'D' are perpendicular

\(\Rightarrow AA'\perp A'C'\)

Let a be the length of each side of the cube in cm.We have :

ABCDA'B'C'D' is a cube,so \(\Delta A'B'C'\) isosceles right at B'

\(\Rightarrow AC^2=2a^2\).

Applying the Pythagoras theorem to the\(\Delta AA'C'\) right at A' ,we have : \(AC'^2=AA'^2+A'C'^2\Leftrightarrow3a^2=9\Leftrightarrow a=\sqrt{3}\)

The total surface area of the cube is : \(6a^2=6\times3=18\) (cm²)