I think question meant : Find the work done in slowly pulling down the system by 8cm. Only then, I am getting the solution you had provided.

 

Let a₁ and a₂ be the extensions in the springs at equilibirum.

 

Start by drawing the FBDS:

 

m₂g = k₂a₂ --- (1)

k₁a₁ = m₁g + k₂a₂ --- (2)

 

a₂ = 0.02m and a₁ = 0.02m.

 

Let x₁ and x₂ be additional elongations causes by pulling m₂ by L = 8cm. The additional forces on  m1 are equal and are in opposite directions.

 

Thus you can infer that k₁x₁ = k₂x₂  --- (3)

Also, use the constraint x₁ + x₂ = L  --- (4)

 

Solve (3) and (4) to obtain the values of x₁ and x₂ as 0.06m and 0.02m respectively.

 

Apply work energy theorem now. Initial and final kinetic energy is 0.

 

W_g + W_p + W_s = 0 --- (5) where W_p is the work done by pulling force.

 

W_s = U₁ - U₂ where U₂ = 1/2 k₁(a₁ + x₁)² + 1/2 k₂(a₂ + x₂)²

 

Put in the values, and you get W_p = 1.2 J.

By the way, if the above is not clear, you might want to refer "Concepts of Physics - H.C Verma). I remember seeing a quite similar problem in it long back.

 

cheers.