Answer
The change of the gravitational pull may be approximated as
$$F_{1}-F_{0} =\frac{G M_{m} m}{\left(R_{M E}-R_{E}\right)^{2}}-\frac{G M_{m} m}{\left(R_{M E}+R_{E}\right)^{2}} $$
$$\approx \frac{G M_{m} m}{R_{M E}^{2}}\left(1+2 \frac{R_{E}}{R_{M E}}\right)-\frac{G M_{m} m}{R_{M E}^{2}}\left(1-2 \frac{R_{E}}{R_{M E}}\right) $$
$$ =\frac{4 G M_{m} R_{E}}{R_{M E}^{3}} $$
On the other hand, your weight, as measured on a scale on Earth, is
$$
F_{g}=m g_{E}=\frac{G M_{E} m}{R_{E}^{2}}
$$
since the moon pulls you "up," the percentage decrease of weight is
$$\frac{F_{1}-F_{0}}{F_{g}}=4\left(\frac{M_{m}}{M_{E}}\right)\left(\frac{R_{E}}{5.98 \times 10^{24} \mathrm{kg}}\right)\left(\frac{6.37 \times 10^{6} \mathrm{m}}{3.82 \times 10^{8} \mathrm{m}}\right)^{3}$$
$$=2.27 \times 10^{-7} \approx\left(2.3 \times 10^{-5}\right) \%$$
Work Step by Step
The change of the gravitational pull may be approximated as
$$F_{1}-F_{0} =\frac{G M_{m} m}{\left(R_{M E}-R_{E}\right)^{2}}-\frac{G M_{m} m}{\left(R_{M E}+R_{E}\right)^{2}} $$
$$\approx \frac{G M_{m} m}{R_{M E}^{2}}\left(1+2 \frac{R_{E}}{R_{M E}}\right)-\frac{G M_{m} m}{R_{M E}^{2}}\left(1-2 \frac{R_{E}}{R_{M E}}\right) $$
$$ =\frac{4 G M_{m} R_{E}}{R_{M E}^{3}} $$
On the other hand, your weight, as measured on a scale on Earth, is
$$
F_{g}=m g_{E}=\frac{G M_{E} m}{R_{E}^{2}}
$$
since the moon pulls you "up," the percentage decrease of weight is
$$\frac{F_{1}-F_{0}}{F_{g}}=4\left(\frac{M_{m}}{M_{E}}\right)\left(\frac{R_{E}}{5.98 \times 10^{24} \mathrm{kg}}\right)\left(\frac{6.37 \times 10^{6} \mathrm{m}}{3.82 \times 10^{8} \mathrm{m}}\right)^{3}$$
$$=2.27 \times 10^{-7} \approx\left(2.3 \times 10^{-5}\right) \%$$
