Solve for the Moon’s case:

As Newton’s law of universal gravitation, The force of gravity between two objects is given by

$$
begin{align*}
F_{M} &= G ~ dfrac{ m_{1} ~ m_{M} }{ r_{M}^{2} } \
end{align*}
$$

Rearrange and solve for the mass of the Moon:

$$
begin{align*}
m_{M} &= dfrac{ F_{M} ~ r_{M}^{2} }{ m_{1} ~ G} \
end{align*}
$$

Solve for the Earth’s case:

As Newton’s law of universal gravitation, The force of gravity between two objects is given by

$$
begin{align*}
F_{E} &= G ~ dfrac{ m_{1} ~ m_{E} }{ r_{E}^{2} } \
end{align*}
$$

Rearrange and solve for the mass of the Moon:

$$
begin{align*}
m_{E} &= dfrac{ F_{E} ~ r_{E}^{2} }{ m_{1} ~ G} \
end{align*}
$$

As the probelm mentions the acceleration due to gravity of the Moon is equal to one-sixth of the acceleration due to the gravity of the Earth $a_{M} = dfrac{1}{6} ~ a_{E}$. Also, the probelm mentions the radius of the Moon is equal to one-quarter of radius of the Earth $r_{M} = dfrac{1}{4} ~ r_{E}$.

divide both equations together, then we get

$$
begin{align*}
dfrac{ m_{M} }{ m_{E} } &= dfrac{ dfrac{ F_{M} ~ r_{M}^{2} }{ m_{1} ~ G} }{ dfrac{ F_{E} ~ r_{E}^{2} }{ m_{1} ~ G} } \
&= dfrac{ dfrac{ m ~ a_{M} ~ r_{M}^{2} }{ m_{1} ~ G} }{ dfrac{ m ~ a_{E} ~ r_{E}^{2} }{ m_{1} ~ G} } \
&= dfrac{ dfrac{ dfrac{1}{6} ~ a_{E} ~ left( dfrac{1}{4} ~ r_{E} right)^{2} }{ m_{1} ~ G} }{ dfrac{ a_{E} ~ r_{E}^{2} }{ m_{1} ~ G} } \
&= dfrac{ dfrac{1}{6} ~ a_{E} ~ left( dfrac{1}{4} ~ r_{E} right)^{2} }{ a_{E} ~ r_{E}^{2} } \
&= dfrac{1}{6} ~ left( dfrac{1}{4} right)^{2} \
&= dfrac{1}{ 96 }
end{align*}
$$

So, the mass of the Moon in terms of the mass of Earth is $dfrac{1}{ 96 } ~ m_{E}$.