32 km above surface and 64 km below the surface. Both.
[tex]g = \frac{G\ M_e}{R_e^2}\\[/tex]
g = acceleration due to gravity at surface of Earth
Re = radius of Earth
Me = Mass of Earth
Acceleration due to gravity " g' " at a height h above the surface is :
[tex]g'=\frac{G\ M_e}{(R+h)^2}=\frac{G\ M_e}{R^2(1+\frac{h}{R})^2}=\frac{g}{(1+\frac{h}{R_e})^2}\\\\g'= \frac{g}{1+\frac{2h}{R}}=g(1-\frac{2h}{R}),\ as\ \frac{h}{R}<<1\\\\1-\frac{g'}{g}=\frac{2h}{R_e}=1\%=0.01\\ \\h=0.01*6400/2=32km\\[/tex]
Hence gravity is less by 1% at a height of 32 km above the surface of Earth.
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Below the surface of Earth:
Let us assume that the density of Earth is uniform below surface. Then
Density of Earth = d = Me / (4πRe³/3)
Let us take a depth of h below the surface of Earth.
Mass M of Earth enclosed within the radius of r = (R-h) contributes towards gravity at the depth h. The volume of Earth in the spherical shell of width h does not contribute to the gravity. This is because all the mass around the shell attracts the mass from all sides and the net effect is zero.
[tex]M = \frac{M_e}{\frac{4}{3} \pi R_e^3}*\frac{4}{3} \pi r^3=\frac{M_er^3}{R_e^3}\\\\g'=\frac{GM}{r^2}=\frac{GM_er}{R_e^2*R_e}=g\frac{r}{R}\\\\\frac{g'}{g}=\frac{R-h}{R}=1-\frac{h}{R}\\\\1-\frac{g'}{g}=\frac{h}{R_e}=1\%=0.01\\\\h=0.01*R_e=64km\\[/tex]
Hence, gravity is 1% less at a depth of 64 km under the surface of Earth.
