Assuming that the Hubble constant H0=74 km/s/mock the critical-density of the universe is 1.0 *10^336-334-31256kg/m^3.
What formula do I use to find:
How much mass is contained in a volume of 1cubic light year?
What is the length of a cube that would enclose 1 earth mass of material?
The formula that is used to evaluate the density is the following,such that:
rho = m/V
m represents the mass of the substance
V reprsents the volume
Hence, if given the density and volume, the mass is evaluated such that:
m = V*rho
On the other hand, one light year measures 8.46*10^47 m^3 and the critical density is 1*10^(-26) (Kg)/(m^3).
m = 8.46*10^47 m^3 * 1*10^(-26) (Kg)/(m^3)
Reducing the similar units of measure yields:
m = 8.46*10^(47-26) Kg
m = 8.46*10^21 Kg
You need to evaluate the length of the cube that encloses one Earth mass, hence, you need to use the volume formula, such that:
V = l^3 => l = root(3)(V)
You need to evaluate the volume of Earth, such that:
V = m/(rho)
The Earth's mass is considered as being 6.4*10^24 Kg.
V = (6.4*10^246.4*10^24 Kg)/(5520*10^3(Kg)/m^3)
V = 1200*10^15 m^3
l = root(3)(1200*10^15 m^3) => l = 1.6*10^6 m