Answer
0.0041
Work Step by Step
Half-life $t_{1/2}=6.0 \,h$
Decay constant $\lambda=\frac{0.693}{t_{1/2}}=\frac{0.693}{6.0\,h}=0.1155\,h^{-1}$
$t=48\,h$
Recall that $\ln(\frac{A}{A_{0}})=-\lambda t$ where $A_{0}$ is the amount of sample at the beginning and $A$ is the amount after time $t$.
$\implies \ln(\frac{A}{A_{0}})=-(0.1155\,h^{-1})(48\,h)=-5.5$
Taking the inverse $\ln$ of both the sides, we have
$\frac{A}{A_{0}}=e^{-5.5}=0.0041$
