What is the total ecliptic surface area of 25,000,000 space asteroids according to these graph from a faraway perspective, i.e the toatl 2D area if we simplify asteroids to faraway disks? It's to know if all the asteroids cover more of the horizon than the moon. Currently I have it as 1-5 times the 2D surface area of the moon(and a lot further away).
What is the figure for the faraway perspective area of all objects from range 50km to .01km?
You need to define what you are trying to calculate. From covering the horizon it sounds like you want to add up the angular diameters of all the asteroids. This would be assuming they are all lined up on the horizon without overlap and asking how much of the circle they cover. The data given is good on the population at each size, so we just need to use the distance to the asteroids. The figure of semi-major axes from [Wikipedia][1] shows the main belt asteroids average about $2.7$ AU $\approx 4\cdot 10^8$ km. We can use this as the distance from earth, though it will vary depending on whether the asteroid is on the same side of the sun as us. The figure you link shows $600k$ asteroids between $1$ and $3$ km, so they occupy about $6\cdot 10^5\frac 2{4\cdot 10^8}\approx 3\cdot 10^{-3}$ radian. You can go through the rest of the data points on the figure and add them up. By comparison, the moon occupies about $\frac {3400}{380000}\approx 9\cdot 10^{-3}$ radian as seen from earth, so we are $1/3$ of the way there with this bin alone.