Almost nowhere. The cavity will be filled with rubble, and while there will be some space between the bigger pieces of rubble, this will not procure enough volume. In
that document I link to, it says the radius for a 100kT explosion at 800m is about 45 meters. Now like I said, the deeper you go, the higher the pressure is. According to
this .pdf document which I think is the correct one, the well drilled by the Deepwater Horizon is of approximately 4.5 km in depth in rock, under 5 km of water. A 100kT explosion that creates a 45m radius cavity at 800m will of course give a smaller cavity at 4.5 km of depth because of the higher pressure at that point. Filled with rubble, the interstitial space does not give a lot of volume for something as big as the Macondo prospect which contains millions of barrels of oil. You would have to place multiple devices in a grid pattern, and therefore drilling more wells to do so. At that point, you have to ask: should we drill two relief wells, or drill multiple wells for nuclear devices which may or may not work (While it is certain that the relief wells will?)
Oh and what's wonderful is that they include mathematical formulae for predicting the size of the cavities that would form in
the document. Let's look at that for an instant shall we?
The formula for predicting the cavity size is this:
R = C (Y^(1/3)/(rho*h)^(0.25))
R is the radius, C is a constant depending on the material, Y is the yield, rho is the density of the material, h is the depth of the explosion. ^ means it's a power (because you can't do small text on the forum.)
The oceanic crust is, on the surface, sedimentary rocks, then basalt, then gabbros and other metamorphic rocks. The biggest part of the oceanic crust is these metamorphic rocks, this is in what the Macondo prospect is, this is what we drilled into to get to it.
A quick search tells us that gabbro density varies from 2.90 to 3.10, let's average this to 3.
Now unfortunately we don't have the C constant in the document for gabbro. But the closest rock in the ones we have is granite, the only other volcanic rock. So for the sake of the argument, let's use it's C constant. I'll average it out to 59.09.
Now our yield is the same, 100kT
Our depth h is 4.5km, or 4500m (Let's exclude the water, even though we shouldn't and treat this as emerged landscape.)
We have: R = 59.09 (100^(1/3)/(3*4500)^0.25)
And so R = 25.4 meters.
So we get a cavity with a diameter of 50 meters or about that, below big oil reservoir, and not all of these 50 meters will be usable because there's a very high chance that it will fill itself with rubble.
Oh and using a higher yield to make the cavity bigger, well I thought about that too. According to
this Wikipedia page, we have the yields for nuclear weapons around the world. However note that we can't use every yield because the device has to fit down into a well bore right? I think the highest would be the W88 nuclear device, and with that yield (500kT) we get a radius of 43 meters. Still not a lot compared to the volume of oil we would want to drain down using the theory you presented. Of course this would also be filled with rubble so it would drain even less.
I don't think this nuclear explosion thing can work at all.
maybe so
