Nobel laureate Richard Feynman once said that if two persons stood at arm's length from each other and each person had p = 1% more electrons than protons, the force of repulsion between them would be enough to lift a “weight” equal to that of the entire Earth. Carry out an order-of-magnitude calculation to substantiate this assertion.

[YoussefLasheen]

Suppose each person has mass $70kg.$ In terms of elementary charges, each person consists of precisely equal numbers of protons and electrons and a needy equal number of neutrons. The electrons comprise very little of the mass, so for each person we find the total number of protons and neutrons, taken together:
$$(70kg)({u\over{1.66×10^{−27}kg​}})=4×10^{28}u$$
of these, nearly on half, $2×10^{28}$,are protons, and 1% of this is $2×10^{26}$, constituting a charge of $$(2×10^{26})(1.60×10^{−19}C)=3×10^7C.$$
Thus, Feyman's force has magnitude
$$F={{ke​q1​q1\over {r2}}​​={{(8.99×109N.m^2/C^2)(3×10^7C)^2}\over {(0.5m)2}​}}∼1026N​$$
where we have used a half-meter arm's length. According to the particle in a gravitational field model, if the Earth were in an externally-produced uniform gravitational field of magnitude 9.80m/s2, it would weigh $$Fg​=mg=(6×10^{24}kg)(10m/s^2)∼10^{26}N.$$
Thus, the forces are of the same order of magnitude.