The half-life of cesium-137 is 30 years. suppose we have a 200-mg sample.

a) To find  the mass after t years:

we will use this formula:

A = Ao / 2^n 

when A =the amount remaining

and Ao = the initial amount

and n = t / t(1/2)

by substitution:

∴ A = 200 mg/ 2^(t/30y)

b) Mass after 90 y :

by  using the previous formula and substitute t by 90 y

A = 200mg/ 2^(90y/30y)

∴ A = 25 mg

C) Time for 1 mg remaining:

when A= Ao/ 2^(t/t(1/2)

so, by substitution:

1 mg = 200 mg / 2^(t/30y)

∴2^(t/30y) = 200 mg  by solving for t

∴ t = 229 y 

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Sorry i dont understand the question can you rephase it?