Answer
21.0 hours
Work Step by Step
Initial activity $A_{0}=53500\,dpm$
Activity after time $t$, $A=10980\,dpm$
Time $t=48.0\,h$
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.
$\implies \ln(\frac{53500\,dpm}{10980\,dpm})=1.5836=\frac{0.693}{t_{1/2}}\times48.0\,h$
Or $t_{1/2}=\frac{0.693\times48.0\,h}{1.5836}=21.0\,h$
