The activity of a nuclide is $15$millicurie. If its decay constant is $0.005$ sec${}^{-1}$, the number of atoms present in it is

Decay constant $=0.005se{c}^{-1}=\lambda$
Activity of nuclei $=15\times 3.7\times 10^{7}dps[\because 1millicurie=3.7\times 10^{7}dps]$
                             $=5.55\times 10^{8}dps$

Now, from differential rate law:
$\frac{dN}{dt}=\lambda N$  [ where N is no of atoms present]

$\Rightarrow \frac{5.55\times 10^8}{0.005}=N\Rightarrow N=11.1\times 10^{10}atoms$