An air-filled parallel-plate capacitor has a capacitance of 1 pf. the plate separation is then doubled and a wax dielectric is inserted, completely filling the space between the plates. as a result, the capacitance becomes 2 pf. the dielectric constant of the wax is:

The equation for the capacitance of a parallel plate capacitor containing a dielectric is: C = (∈oΚA)/d, where, C is the capacitance, ∈o is the permittivity of space, Κ is the dielectric constant, A is the area of the plate, and d is the distance between the plates. Thus, when the plate separation is doubled, the capacitance is halved and since the capacitance increased from 1 pf to 2 pf, it can be concluded that the dielectric constant of the wax is equal to 4. 

Edit the post so i can see the letters for each answer, and increase the points required so i'd be able to answer it.