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Answer 1

13 half-lives have passed

Further explanation

General formulas used in decay:  

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}$

T = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

t1/2 = 4 days

Nt=18 mg (0.01% of the original isotope)

18 mg (Nt) = 0.01% No

No = the original isotope :

$\tt No=\dfrac{100}{0.01}\times 18~mg=180,000$

The duration of decay  (T) :

$\tt \dfrac{Nt}{No}=\dfrac{1}{2}^{T/4}\rightarrow Nt=0.01\%No$

$\tt 0.01\%=\dfrac{1}{2}^{T/4}\\\\10^{-4}=\dfrac{1}{2}^{T/4}\\\\(\dfrac{1}{2})^{13}=\dfrac{1}{2}^{T/4}\\\\13=T/4\rightarrow T=52~days$

Half-lives passed :

$\tt \dfrac{52}{4}=13~half-lives$