Solution

Suppose the potential energy of an isolated face of the tetrahedron is $U_1$. Then, the potential energy of two faces of the tetrahedron pressed together is $4U_1$, because the charge is doubled in the same space. Let the potential energy of two faces of the tetrahedron making a dihedral angle with each other be $2U_1+U_2$. Then the work done in pressing the two faces together is $A=4U_1-(2U_1+U_2)=2U_1-U_2$.

Originally, the potential energy of the tetrahedron is $4U_1+6U_2$, because there are $6$ possible pairs out of $4$ faces. When the four faces are pressed together, the potential energy is $16U_1$, because the charge is quadrupled in the same space. Thus, the work done in collapsing the tetrahedron is $16U_1-(4U_1+6U_2)=12U_1-6U_2=6A$.