Answer
$2.18$
Work Step by Step
$$ \frac{F_\mathrm{S}}{F_\mathrm{M}} = \frac{\frac{GM_\mathrm{S}M_\mathrm{m}}{R_\mathrm{orb-E}^2}}{\frac{GM_\mathrm{E}M_\mathrm{m}}{R_\mathrm{orb-m}^2}} = \frac{M_\mathrm{S}R_\mathrm{orb-m}^2}{M_\mathrm{E}R_\mathrm{orb-E}^2} = $$
$$ \frac{(1.99 \times 10^{30} \, \mathrm{kg})(3.84 \times 10^{8} \, \mathrm{m})^2}{(5.97 \times 10^{24} \, \mathrm{kg})(1.50 \times 10^{11} \, \mathrm{m})^2} = 2.18 $$
