Answer
34.6 days
Work Step by Step
Initial activity $A_{0}=8540\,dpm$
Activity after time $t$, $A=6990\,dpm$
Time $t=10.0\,d$
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.
$\implies \ln(\frac{8540\,dpm}{6990\,dpm})=0.2003=\frac{0.693}{t_{1/2}}\times10.0\,d$
Or $t_{1/2}=\frac{0.693\times10.0\,d}{0.2003}=34.6\,d$
