(B) The ratio of the radius of the electron's orbit to the radius of the nucleus is $(10^{-10} \; m) / (10^{-15} \; m) = 10^{5}$. This means the radius of the electron's orbit is $10^{5}$ times larger than the radius of the nucleus.
If the solar system had the same proportions,the radius of the earth's orbit would be $10^{5}$ times the radius of the sun.
Calculated radius $= 10^{5} \times (7 \times 10^{8} \; m) = 7 \times 10^{13} \; m$.
The actual orbital radius of the earth is $1.5 \times 10^{11} \; m$.
Since $7 \times 10^{13} \; m > 1.5 \times 10^{11} \; m$,the earth would be much farther away from the sun than it actually is.
This implies that an atom contains a much greater fraction of empty space than our solar system does.