Density of nucleus is of the order of

N = N₀e^–λt

->t = λN = N₀/e

So number of nuclei decayed = N₀ – N

= N₀ (1- 1/e) 

Density of nuclei is order of 10¹⁷ kg m^–3.

Number of half lives of Y = 3

Number of half lives of X = 6 [As half life of X is half of that of Y].

$\frac{N_1}{2^6}=\frac{N_2}{2^3}\Rightarrow \frac{N_1}{N_2}=8$

According to Nuclear activity, we can write

$N_0\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{2}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{4}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{8}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{16}=\left(0.0625\right)N_0$         

Time required = 4 × $t_{\frac{1}{2}}=20yrs$