Answer
The person's weight decreases by $~~0.30~N$
Work Step by Step
Note that a person's weight is proportional to $\frac{1}{r^2}$, where $r$ is the distance between the person and the center of the Earth.
The radius of the Earth is $r_E = 6.37\times 10^6~m$
We can find the distance to the center of the Earth when a person goes to the top of the building:
$r' = 6.37\times 10^6~m + 1600~m = 6.3716\times 10^6~m$
We can find the factor by which to multiply the person's weight on the Earth's surface:
$\frac{1/(r')^2}{1/r_E^2} = \frac{r_E^2}{(r')^2} = \frac{(6.37\times 10^6~m)^2}{(6.3716\times 10^6~m)^2} = 0.9995$
We can find the person's weight:
$weight = (0.9995)(600~N) = 599.70~N$
The person's weight decreases by $~~0.30~N$
