Gravitational force on a body of mass $m$ at height $h$ due to earth,
$$
F=\frac{G M_e m}{(R+h)^2}
$$
where, $M_e$ is mass of the earth.
$$
\because \text { Density of earth, } \rho=\frac{\text { mass of earth }\left(M_e\right)}{\text { volume of earth }(V)}
$$
or $M_e=\rho \cdot V=\rho\left(\frac{4}{3} \pi R^3\right)$
$$
\begin{gathered}
\therefore \quad F=\frac{G\left(\frac{4}{3} \pi R^3\right) \rho m}{(R+h)^2} \\
\text { or } \quad F=\frac{G\left(\frac{4}{3} \pi R\right) \rho m}{\left(1+\frac{h}{R}\right)^2} \\
\text { or } \quad F=\frac{4}{3} \pi G R \rho m\left(1+\frac{h}{R}\right)^{-2}
\end{gathered}
$$
or
$$
F=\frac{4}{3} \pi G R \rho m\left(1+\frac{h}{R}\right)^{-2}
$$

By using Binomial expansion,
$$
F=\frac{4}{3} \pi G R \rho m\left(1-\frac{2 h}{R}\right)
$$
$\left[\because \quad(1+x)^n=1+n x+\frac{n(n-1)}{2 !} \cdot x^2+\ldots\right.$ and neglecting higher terms.]

Difference in weight $=\frac{4}{3} \pi G R \rho m-\frac{4}{3} \pi G \rho m R\left(\frac{2 h}{R}\right)$
Hence, error in weight $=\frac{8}{3} \pi \rho G m h$.