Answer:
A0=3kg
t1/2=67y
λ=ln2/t1/2
λ=0.6931/67
λ=0.0103447761-> 0.0103
Final Amount=Aoe^-λ×t
Final Amount=A0e^-0.0103447761×t
You can put the time that you want to find how much of the material has been left after a certain amount of time
as an example
if t=143years
Final Amount=3×e^-0.0103447761×143
Final Amount=3×e^−1.479302982
Final Amount=0.6833892338
A0 is the amount of material that we have before it goes into radioactive decay