density of water * V(submerged) * g = density of the object * V(object) * g. This rearranges to:
V(submerged) / V(object) = density of the object / density of water. We're told V(submerged) / V(object) = 1/2. So the density of the object = 500 kg / m^3 because the density of water = 1000 kg / m^3.
The equation above holds for the second setup in Hg. The specific gravity of Hg is given as 13, which means its density = 13000 kg / m^3.
density of the object / density of Hg = 500 kg / m^3 / 13000 kg / m^3 = 1/26 = V(submerged) / V(object) which is closest to answer A (they should have given the actual answer).
This setup is really common. I would recommend memorizing the initial Fb = mg step because it applies to almost every buoyancy problem that I see.
Okay, quick way to the answer. 1/2 of the volume of the cork is submerged in water. Now in mercury, which is 13 times as dense as water, how much is submerged? Well, realize that how much of the object is submerged depends on the density of the object relative to the liquid it's floating in. Say the density is like 0.7 g/mL (the math doesn't necessarily work out - I just made up a number). You know that it will float in water because it is less dense than 1.0 g/mL. Now put it in something that's 13 times as dense, or 13.0 g/mL. Okay, now the differential is even larger! So if it floats okay in water, it will float amazingly well in the denser liquid!
Now look at the answer choices. If it will float very well, very little of it will be submerged. So that rules out B and C. D can then be ruled out because it's only a little smaller than 1/2 - you can't have a 13-fold change in density of the liquid and only a minute difference in submersion.