Answer
c) 0.935 dps
Work Step by Step
Original activity $A_{0}=15.2\,dps$
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{5715\,y}=1.2126\times10^{-4}\,y^{-1}$
Time $t=23000\,y$
$\ln(\frac{A_{0}}{A})=kt$ where $A$ is the current activity.
$\implies \ln(\frac{15.2\,dps}{A})=1.2126\times10^{-4}\,y^{-1}\times23000\,y=2.789$
Taking the inverse $\ln$ of both sides, we have
$\frac{15.2\,dps}{A}=e^{2.789}=16.26$
Or $A= \frac{15.2\,dps}{16.26}=0.935\,dps$
