Let $M_E$ be the mass of the Earth and m be an object on the surface of the Earth. Let $R_E$ be the radius of the Earth. According to the universal law of gravitation, weight $W_E$ of the object on the surface of the Earth is given by,

$W_{E}=\frac{G M_{E} m}{R_{E}^{2}}$

Let $M_M$ and $R_M$ be the mass and radius of the moon. Then, according to the universal law of gravitation, weight $W_M$ of the object on the surface of the moon is given by:

$W_{M}=\frac{G M_{M} m}{R_{M}^{2}}$

Now,

$\Rightarrow \frac{W_{M}}{W_{E}}=\frac{M_{M} R_{E}^{2}}{M_{E} R_{M}^{2}}$

Where,

$M_{E}=5.98 \times 10^{24}\, kg$

$M_{M}=7.36 \times 10^{22} \,kg$

$R_{E}=6.4 \times 10^{6} \,m$

$R_{M}=1.74 \times 10^{6} \,m$

$\Rightarrow \frac{W_{M}}{W_{E}}=\frac{7.36 \times 10^{22} \times\left(6.4 \times 10^{6}\right)^{2}}{5.98 \times 10^{24} \times\left(1.74 \times 10^{6}\right)^{2}}=0.165 \approx \frac{1}{6}$

Therefore, weight of an object on the moon is $\frac {1} {6}^{th}$ of its weight on the Earth.