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Answer: 90.2 days

Explanation:-

Radioactive decay follows first order kinetics.

Half-life of Iron-59 = 45.1 days

[tex]\lambda =\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{45.1}= 0.0153 days^{-1}[/tex]

[tex]N=N_o\times e^{-\lambda t}[/tex]

N = amount left after time t= [tex]\frac{25}{100}\times a=0.25a[/tex]

[tex]N_0[/tex] = initial amount  = a

[tex]\lambda[/tex] = rate constant= [tex]0.015 days^{-1}[/tex]

t= time  = ?

[tex]0.25a=a\times e^{- 0.0153 years^{-1}\times t days}[/tex]

[tex]t=90.2days[/tex]

Thus the iron sample is 90.2 years old.

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