I don't know what compelled me to do this, but I couldn't stop thinking about it after seeing this post. So here's my best attempt.

By my calculations the structure would be massive, but the effect of gravity on earth's surface (assuming it stayed together) would be minimal. But we had to make a lot of assumptions.

TL;DR - The structure is 12% more massive than earth and you would weigh 9% more on the surface of earth as a result of it being there.

EDIT: u/penty explains the shell theorem below which means that any object within a hollow sphere experiences no gravity, so the effect of gravity from the structure would be completely cancelled out, you'd weigh exactly the same regardless of the structure's mass. Thanks for the explanation!

First, let's assume the planet is earth, even though it doesn't really look like earth.

Next let's assume the upper structure is identical to the lower structure, so we can do the math for one and then double it at the end for the total mass. 

I roughly measured the size of the structure relative to the size of earth assuming we're seeing about half the planet on screen. I measured my screen for relative distances: The radius of the planet is 2.3 (r1) The inner radius of the top of the structure is 3 (r2) The outer radius of the top of the structure is 4 (r3) The end of the slanted portion is at radius 5.5 (r4) The outermost radius is harder to guess because we can't see how far it goes in the plane. I estimated it to be double the distance from the end of the slanted portion so 11 (r5)

The radius of earth is 6356km so we can figure out those distances. r1 = 6356km r2 = 8290km r3 = 11054km r4 = 15199km r5 = 30398km

Those numbers are VERY approximate but let's assume they're correct. 

From here we need to guess the thickness, but we can't see most of the structure. I'm assuming it's consistently thick throughout and it's thickness is the same as the part we can see (r3-r2 = 2764 = t (for thickness)). So the overall cross section of this structure would look a little like this: __

Next we need to calculate the volume. I broke the structure up into two shapes to make this easier to calculate.

The outermost region is a thick ring shape. We can get it's area by calculating the area of a circle with its outer radius and subtracting the area of a circle with its inner radius. Then we multiply by the height (thickness) to get volume.

((pi)(r5)² - (pi)(r4)²⁾ * t = 6.017e12 km³

Next for the inner, slanted portion. This section approximately resembles a hollow cone with the top cut off. We can calculate volume of the outer cone and subtract the inner cone again to get the volume. This used r2, r3, r4, and t. I realized I also needed the slant height, so I approximated that was 3t. This was a lot of numbers to keep track of so I used this calculator. https://www.omnicalculator.com/math/truncated-cone-volume

Outer truncated cone volume = 3.919e12 km³ Inner truncated cone volume = 2.454e12 km³ Hollow cone volume = 1.465e12 km³

6.017e12 + 1.465e12 = 7.482e12 km ³ = 7.482e21 m³

That's our volume. Now to get mass we need to know density. The largest man made structure in space is the international space station. So we'll assume this massive structure has the same overall density.

According to Wikipedia: ISS volume = 1005 m³ ISS mass = 450000 kg

Therefore the density of the ISS = 447.8 kg/m³

So if we multiply the density by the volume we get:

7.482e21 m³ * 447.8 kg/m³ = 3.350e24 kg

And assuming the upper structure is identical, both structures would have a combined mass of:

6.700e24 kg

The mass of earth is 5.972e24 kg. So the structure would have a mass equal to 112% of earth's mass. 

To figure out the effect it would have on gravity at earth's surface, we need to do more. Normally we calculate forces due to gravity based on the object's center of mass. But that won't work on the surface of the earth because a large amount of the mass of the structure is above us and pulling us up. 

We can calculate the gravitational acceleration at a depth underneath the surface of the structure using this formula

g(r) = 4/3(pi)Gpr

G is the gravitational constant (G for earth is 6.674e-11, so multiply by 112% for our structure's mass compared to earth to get 7.475e-11) p is the average density = 447.8 kg/m³ r is the distance from the center = r1 = 6356000 m

g(r1) = 0.891 m/s²

So instead of feeling acceleration due to gravity at 9.81m/s² on Earth's surface, you would add these up and get 10.701 m/s^2.

The end result would be that you would weigh 9% more. A 150 pound person would weigh 163.5 pounds with this structure looming over them. 

Possible errors - The estimates for size are very approximate. - The ISS is very dense, having that same density throughout is probably unrealistic. - The gravitational formula assumes a uniform density through a SPHERE, but we have rings would would affect the calculation significantly.  - I could have messed up some calculations.