Answer
$145 \text{ pounds}$
Work Step by Step
The variation model described by the problem is $
w=\dfrac{k}{d^2}
,$ where $w$ is the weight of an object on or above the surface of the Earth and $d$ is the distance between the object and the Earth's center.
Substituting the known values in the variation model above results to
\begin{array}{l}\require{cancel}
160=\dfrac{k}{4000^2}
\\\\
k=160(4000^2)
\\\\
k=160(16000000)
\\\\
k=2,560,000,000
.\end{array}
Therefore, the variation equation is
\begin{array}{l}\require{cancel}
w=\dfrac{2,560,000,000}{d^2}
.\end{array}
Using the variation equation above, then
\begin{array}{l}\require{cancel}
w=\dfrac{2,560,000,000}{4200^2}
\\\\
w=\dfrac{2,560,000,000}{17,640,000}
\\\\
w\approx145.12
.\end{array}
Rounded to the nearest whole pound, the weight is $
145 \text{ pounds}
.$
