Here, each side of the cube L=4 cm
Area of the face over which the force is applied,
`A=L^(2)=16` sq. cm
Displacement, `DeltaL=0.1` mm=0.01cm
Force applied, F=8 kN=`8000xx10^(5)=8xx10^(8)` dyne
As, `eta=(FL)/(ADeltaL)`
Shear modulus of the cube,
`eta=(8xx10^(8)xx4)/(16xx0.01)=2xx10^(10)` dyne/sq. cm