Answer
(a) 27.4 years
(b) 182 years
Work Step by Step
(a) Initial activity $A_{0}=1.0\times10^{3}\,dpm$
Activity after time $t$, $A=975\,dpm$
Time $t=1\,y$
Recall that $\ln(\frac{A_{0}}{A})=kt=\frac{0.693}{t_{1/2}}\times t$ where $k$ is the decay constant and $t_{1/2}$ is the half-life.
$\implies \ln(\frac{1.0\times10^{3}\,dpm}{975\,dpm})=0.025318=\frac{0.693}{t_{1/2}}\times1\,y$
Or $t_{1/2}=\frac{0.693\times1\,y}{0.025318}=27.4\,y$
(b) $\ln(\frac{100}{1})=4.605=\frac{0.693}{27.4\,y}\times(t)$
$\implies t=\frac{4.605\times27.4\,y}{0.693}=182\,y$
