The question is incomplete. The complete question is :
The base of an aluminum block, which is fixed in place, measures 90 cm by 90 cm, and the height of the block is 60 cm. A force, applied to the upper face and parallel to it, produces a shear strain of 0.0060. The shear modulus of aluminum is
. What is the displacement of the upper face in the direction of the applied force?
Solution :
The relation between shear modulus, shear stress and strain,

Shear stress = shear modulus (S) x shear strain

Pa
Pa

The length represents the distance between the fixed in place portion and where the force is being applied.
Therefore,

= 0.006 x 60 cm
= 0.360 cm
= 3.6 mm
Thus, the displacement of the upper face is 3.6 mm in the direction of the applied force.