If the observable universe were compressed into a super massive black hole, how big would it be?
Calculation
First I went here, getting the number 3.14×10^54 kg.
Under "the size of a black hole" I understand the Schwarzschild radius. Which is $r=\frac{2 G M}{c^2}$.
Then I just did the math:
http://www.wolframalpha.com/input/?i=2G%2A(3.14E54+kg)%2Fc%5E2
Obtaining ~5 diameters of observable Universe...
The Problem
I felt really uneasy about the result. Indeed: I took the mass of observable Universe and obtained 5 times radius of the observable Universe. But how on Earth the observable Universe is smaller than the black hole you can make from it?!
To elaborate, let's also calculate how the "density" of the black hole depends on its mass:
$\rho(M) = \frac{3M}{4\pi r^3}=\frac{3Mc^6}{4\pi\cdot 8 G^3 M^3} = \frac{3 c^6}{32\pi G^3}\cdot\frac{1}{M^2}$
We got that with the growth of mass of the black hole its "density" decreases. At really high masses it gets so disperse that it is even gets smaller than the density of the Universe as we see it.
Interpretation and the answer
Now, how do I understand this. The black hole is "a region of space from which nothing, not even light, can escape". So, in order to make sense of the question we got to imagine some (presumably, flat) empty space. And then we are putting all the matter we gathered into it. It turns out, that even if we do not "compress" this stuff (we can even make it $\simeq5^3=125$ times more disperse), the gravitation attraction of the system would be strong enough for "nothing, not even light, could escape" from the system.
So, as far as I understand the situation now, answer to the question is: aliens don't need to do anything with the Universe in order to make a black hole (in a sense) of it.
About 73% of the energy of the Universe is stored in dark energy - the cosmological constant, most likely - which has a negative pressure numerically equal to the energy density. Because it is a cosmological "constant", this portion of the mass of the Universe cannot be really compressed. So it is a problematic idea to include the dark energy into the "mass that you want to compress".
Only dark matter and visible matter - whose pressure is nearly zero - can really be "compressed". So that would be about $1\times 10^{54}$ for the visible Universe. $2GM/c^2$ for this mass $M$ produces about 150 billion light years which is about 3 times larger than the radius of the observable Universe.
We obtained the result that even if we only count the "particulate" component of the mass of the Universe, we find out that the Universe is actually smaller than the black hole. It is no contradiction because the Universe is not a static system. It would be impossible for the Universe to be kept this small. However, the galaxies are receding from each other which prevents the formation of a nearby event horizon. Instead, the closest horizon in the reality is the "cosmic horizon" - how far we can actually see because of the finite speed of light and finite age of the Universe. It depends on the observer but effectively makes everything beyond this horizon unphysical.
Our Universe, dominated by the dark energy, is already rather close to an empty de Sitter space which is, from many viewpoints, analogous to a black hole except that the interior of the visible part of the de Sitter space is analogous to the exterior of a normal black hole, and the analogy of the interior of a black hole is everything that is behind the cosmic horizon - where we don't see. It is misleading to create the analogy with the static black holes directly because our Universe is not static in the normal cosmological coordinates.