Answer
$ f(80)=157.49$ kg. This is unsafe.
Work Step by Step
In the year 2066, $x$ will be $80$.
$ f(80)=1000(0.5)^{80/30}\approx 157.49$
Which is more than the amount considered to be safe (100 kg).
Thus, Chernobyl will not be safe by 2066.
There will be $157.49$ kg of cesium-137 remaining in Chernobyl's atmosphere.
