As you know, Archimedes' principle states that the buoyant force experienced is equal to the weight of the displaced fluid.
In the case of your incompressible water, the buoyant force from the water experienced by the block of wood is independent of the air pressure as long as the block is immersed to the same extent: any excess pressure to the entire system would act equally on the top and bottom of the block and cancel out.
But wait - there's more.
The question did NOT say to ignore the compressibility of the AIR. If you increase the density of the air (by raising the pressure), the weight of the displaced air increases. This in turn means that the block experiences a greater buoyant force due to the air, and that it will therefore rise (a little bit) in the water in order to find a new "neutral buoyancy" position.
This is counterintuitive: you raise the pressure and the block rises up out of the water. But I'm pretty sure that's correct. The answer should have been (B).
If we ignore the compressibility for a moment and look just at the pressure of the air, we find that it is "all around us"; not only that, but (barring the effect of gravity) it is the same everywhere. As David Hammen pointed out in his answer, sometimes, in the real world, you have to know what to ignore. If you pick up a ball, you don't have to think of it as a bunch of atoms with electrons forming bonds, obeying Schroedinger's equation, ... you can just think of it as a "ball".
In the same way, for most practical purposes we can think of air as something that
- fills every space in our experiment (unless we stop it)
- has very low density
- has a pressure of about 1 kg/cm3
- will provide a small amount of drag to objects moving through it
Most of the time that is all you need to know about air. In the case of your block, if you have a pressure of 1 bar (normal atmospheric pressure) above the block, that same pressure is exerted on the surface of the water - and so at the very top of the water level, the pressure is also 1 bar. As you go deeper in the water, the pressure is even greater, because of the weight of the water "above" the point where you are measuring.
If you increase the atmospheric pressure, the pressure inside the water increases by the same amount. This means that the difference in pressure between top and bottom of the object is independent of the pressure of the air. And the difference is what gives rise to the buoyancy.
Let's look at this picture:
If the block has surface area (top and bottom) $A$ and the atmospheric pressure is $P$, then the force pushing down on the top is
$$F_1 = P\cdot A$$
The pressure at the bottom of the block $P_2 = P + \rho \cdot g \cdot h$ and the force on the bottom is
$$F_2 = P_2 \cdot A = \left(P + \rho \cdot g \cdot h\right) \; A$$
The difference between these forces is what is experiences as buoyancy, and has to equal the weight of the block:
$$m\cdot g = F_2 - F_1 = \rho \cdot g \cdot h \cdot A$$
As you can see, the term $P$ canceled out.
This is true regardless of the shape of the block: it is sufficient to think of the block as made up of many smaller blocks, each of a regular shape, and add up all the forces due to each "blocklet". For each, the $P$ term will cancel out.
Now with some practice you will "know" when you can ignore pressure - until you do, you can (and should) do the more rigorous analysis to convince yourself that you can ignore it.
Knowing what not to do takes a lifetime of learning: but it can make life so much easier...
