It is actually two separate binaries--one consisting of a red giant 100 times as wide, one-third as massive and 1,000 times as bright as our sun orbited by a yellow-white dwarf 130% as wide, 170% as massive and six times as bright as our sun; another consisting of two yellow dwarves, each one 105% as wide, 110% as massive and 126% as bright as our sun. This alternate Earth orbits the latter binary
A yellow-white dwarf should be equal to a spectral class F star, I guess. A star 130% as wide as the Sun would have 219.7 percent of the volume of the Sun, and if it had the same density it would have 219.7 percent of the mass of the Sun, not 170 percent.
The yellow dwarf stars with 105 percent the diameter of the Sun would have 115.76 the volume of the Sun, and thus its mass, instead of 110 percent the mass of the Sun.
You should check the physicial characteristics of various spectral types of stars in user177107's answer to the question:
The chart claims that:
A G5V star would have 0.98 of the Sun's mass, 0.982 of the Sun's radius, and 0.887 of the Sun's luminosity.
A G2V star would have 1.02 of the Sun's mass, 1.01 of the Sun's radius, and 1.014 of the Sun's luminosity.
A F8V star would 1.18 have of the Sun's mass, 1.25 of the Sun's radius, and of 2.031 the Sun's luminosity.
A F5V star would have 1.33 of the Sun's mass, 1.46 of the Sun's radius, and 3.434 of the Sun's luminosity.
A F2V star would have 1.44 of the Sun's mass, 1.61 of the Sun's radius, and 9.3 of the Sun's luminosity.
A A8V star would have 1.67 of the Sun's mass, 1.81 of the Sun's radius, and 14.392 of the Sun's luminosity.
I don't think that your figures for the diameters and massess and luminosities of the stars are quite correct.
A red giant star is one which has completed its main sequence period. In a star system where the stars are the same age, the most massive star will completed it smain sequence priod first and become a red giant star first. So I fail to see how the red giant could have one third the mass of the sun, and if it did it would orbit the yellow-white dwarf instead of being orbited by it.
A separation of 10 AU between the two pairs of stars seems insufficient to protect the "Alternate Earth" from a runaway greenhosue effect as the red giant becomes more luminous. A planet 10 AU from the Sun would recieve only 1 percent of the radiation from the Sun as a planet 1 AU from the Sun would receive. Yousay the red giant is 1,000 times as luminous as the Sun.
So the "Alternate Earth" 10 AU from the giant star would receive 10 times as much radiation from from the red giant as Earth receives from the Sun. That should be enough to cause a runaway greenhouse effect, destroying all life on the planet.
Remember that luminosity decreaes with the square of the distance. So if the absolute luminosity of a star in your system equals more than the luminosity of the Sun divided by the square of the distance in AU to a planet in that system, that planet will receive more radiation from that star than Earth gets from the Sun, and will thus be hotter than Earth.
Why do you write:
10 AUs, safe from the intense flares of the yellow-white dwarf.
As far as I know, some red dwarf stars o are the mainsequance stars with the most intense flares, which is why those red dwarf stars are known as flare stars.