Answer
The multiple of the energy needed to escape from Earth which gives the energy needed to escape from the Moon is $~~0.0451$
Work Step by Step
We can write a general expression for the escape speed:
$v = \sqrt{\frac{2GM}{R}}$
We can write a general expression for the kinetic energy required to escape:
$K = \frac{1}{2}mv^2$
$K = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2$
$K = \frac{GMm}{R}$
We can find the ration $\frac{K_M}{K_E}$, which is the ratio of the kinetic energy required to escape from the moon to the kinetic energy required to escape from the Earth:
$\frac{K_M}{K_E}= \frac{\frac{GM_Mm}{R_M}}{\frac{GM_Em}{R_E}}$
$\frac{K_M}{K_E}= \frac{M_M~R_E}{M_E~R_M}$
$\frac{K_M}{K_E}= \frac{(7.36\times 10^{22}~kg)~(6.37\times 10^6~m)}{(5.98\times 10^{24}~kg)~(1.74\times 10^6~m)}$
$\frac{K_M}{K_E}= 0.0451$
The multiple of the energy needed to escape from Earth which gives the energy needed to escape from the Moon is $~~0.0451$
