[tex]\bf \begin{array}{llllll}
\textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\
\textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\
y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x}
\\
&&y=\cfrac{{{ k}}}{x}
\end{array}[/tex]
with "k" being the "constant of variation"
so. hmm in this case is "w" inversely varying to the square of the distance or d²
thus [tex]\bf w=\cfrac{k}{d^2}\qquad
\begin{cases}
d=3978\\
w=129
\end{cases}\implies 129=\cfrac{k}{3978^2}[/tex]
solve for "k", to see what the constant of variation is
and plug it back in [tex]\bf w=\cfrac{k}{d^2}[/tex]
now.. .to find her weight when she's 144miles above the surface, well
that's just [tex]\bf w=\cfrac{k}{(3978+144)^2}[/tex]
and since you already know what "k" is, just divide and simplify away, to get "w"