What is the 'weight' (not mass!) of the Earth? I am asking what might be the weight (not mass!) of the earth. My weight is the amount of force exerted on me by Earth (say 600N). So, weight of earth should be amount of force exerted on Earth by me (600N)? And it also seems to be changing from person to person. Where am I wrong?
 A: I suppose the closest thing would be the pressure at the core, but that's not a weight as such.
For comparison an average human standing on the surface of Earth exerts a pressure of about 16 psi or 110 kPa and the atmospheric pressure is about 100 Kpa.
The pressure at the core is thought to be about 300 GPa.  So about 3 million times more at the core that the effect of an average human on the surface.
And to put that in perspective, the average human adult has a mass of around $70 kg$ and the Earth has a mass of around $6\times 10^{24} kg$, so the human is sort of punching well above it's weight, so to speak. :-)
A: Weight is a force, with which one object attracts another via gravity. Therefore weight is relative to the object. If your weight on the Earth is 600N, then your weight on the Moon would be only 100N. And if your weight on the Earth is 600N while mine is 1000N, then the weight of the Earth on you would be 600N, while the weight of Earth on me would be 1000N at the same time. So clearly, as others have pointed out, measuring the weight of Earth has no practical meaning and therefore is not a useful concept. 
A: We can, if we like, work out the "weight" of Earth due to Earth's own gravity, which is far greater than the gravity we individually exert on it. Let's assume constant density (the truth is more complicated, but will give an answer within an order of magnitude of what we get this way).
Say Earth has mass $M$ and radius $R$, so its surface gravity is $g:=GMR^{-2}$. (Obviously, we expect an answer comparable to $Mg$.) However, at a distance $r<R$ from Earth's centre the gravitational field strength is $G\frac{Mr^3/R^3}{r^2}=g\frac{r}{R}$. The spherical shell of radius $r$, thickness $dr$ has volume $4\pi r^2 dr$, which is a fraction $\frac{3r^2}{R^3}dr$ of Earth's volume $\frac{4}{3}\pi R^3$. So this mass-$\frac{3Mr^2}{R^3}dr$ shell feels a gravitational field strength $\frac{gr}{R}$, and has weight $\frac{gr}{R}\cdot\frac{3Mr^2}{R^3}dr=\frac{3Mgr^3}{R^4}dr$. Now we just integrate:$$\int_0^R \frac{3Mgr^3}{R^4}dr=\frac{3Mg}{R^4}\frac{R^4}{4}=\frac{3}{4}Mg.$$As already mentioned, a non-uniform density gets a different coefficient for $Mg$. For what it's worth, $\frac{3}{4}Mg=\frac{3GM^2}{4R^2}$.
A: I think that you are right in your analysis.
The weight of an object on the Earth is short for the attractive force on an object due to the object being in the gravitational field of the Earth.
Ignoring the rotation of the Earth it is the reading on the bathroom scales on which the object is placed.
Now look at the situation of the object creating the gravitational field and the Earth being in that gravitational field.
You can then think of the bathroom scales as measuring that force but I do not think that this force is generally called the weight of the Earth.
The reason that you do not usually worry about the force on the Earth due to the object is because that force does not change the motion of the Earth by very much because the Earth is so massive as compared to the object.
Update as a result of a comment made about the following statements made in my answer:

it [the force on the object due to the gravitational attraction of the Earth] is the reading on the bathroom scales on which the object is placed.
You can then think of the bathroom scales as measuring that force [the force on the Earth due to the gravitational attraction of the object].

The reason that bathroom scales show a reading which is equal to the force on the object due to the gravitational attraction of the Earth is as follows.
In static equilibrium on the Earth the object has two equal in magnitude and opposite in direction forces acing on it (Newton's second law) - the force on the object due to the gravitational attraction of the Earth and the force on the object due to the spring in the bathroom scales.
The Earth has two equal in magnitude and opposite in direction forces acing on it (Newton's second law) - the force on the Earth due to the gravitational attraction of the object and the force on the Earth due to the spring in the bathroom scales.
The spring has two equal in magnitude and opposite in direction forces acing on it (Newton's second law) - the force on the spring due to the Earth and the force on the spring due to the object.
These two forces acting on the spring cause a change in the spring which is shown as a reading on the bathroom scales equal to the magnitude of those forces acting on the spring.
The Newton's third law pairs of forces which make them equal in magnitude and opposite in direction are:
The force on the object due to the gravitational attraction of the Earth and the force on the Earth due to the gravitational attraction of the object.
The force at one end of the spring due to the Earth and the force on the Earth due to that end of the spring.
The force at the other end of the spring due to the object and the force on the object due to that other end of the spring.
Since the magnitude of all the forces is the same one can say that the reading on the bathroom scales is equal to the force on the object due to the gravitational attraction of the Earth and also the force on the Earth due to the gravitational attraction of the object.
The force on the object due to the gravitational attraction of the Earth is called the weight of the object but the force on the Earth due to the gravitational attraction of the object is generally not called the weight of the Earth.
A: It seems you are trying to apply weight inconsistently.
We generally define weight as "force on an object due to gravity."
You cannot really define weight of an individual object.  When defining weight we usually are implicitly talking about the objects weight on Earth.  We can also find the weight of objects on the moon for example.  Objects with the same mass will obviously have different weight in different gravity.
This is because gravity is an attractive force between two objects.  The question of "What is the weight of Earth?" has no obvious answer, because generally when we don't specify where, we mean on Earth.  The question would then read "What is the weight of Earth on Earth?", which is obviously nonsense.
In your example, you could ask the question "What is the Earth's weight on me?", and then it would be 600 N.  The weight of the Earth for someone heavier would be more.
It's not really standard to talk about the Earth's weight on small objects either, which is why it may seem odd to consider that.
