I'm having a bit of trouble with this problem, you can neglect any effect at the boundaries.
A parallel plate capacitor, whose plates have a distance 2d, is charged and then isolated. Keeping the charge on each plate constant, we introduce, parallel to the plates, a slab of dielectric of thickness d, and relative dielectric constant k. We then extract the insulating slab, and introduce a conducting slab, also of thickness d. Knowing that, in order to extract the insulating slab, we need to do a work W1, and to extract the conducting slab a work W2 = 3W1, compute k.
So since Q is constant, initially the capacitor has capacitance;
C0 = Q/V0 = A(ε0)/2d
after the insulating slab of dielectric is added we compute;
C = kC0=εA/2d where ε = k(ε0)
and when the conducting slab is introduced, the initial capacitor essentially becomes two such that the inverse of the total capacitance:
1/Ct = 1/C1 + 1/C2, where C1 = C2 = A(ε0)/(d/2), therefore Ct = A(ε0)/d
I know this isn't enough to solve the problem, and that the electrostatic energy stored in the capacitor U0 = C0(V0^2)/2 = QV0/2 = Q^2/2C0.
What is W1? Is it the sum of the changes in energy when the slabs are inserted and removed? and how can I use this to find k?