Answer
18 hours.
Work Step by Step
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{6.0\,h}=0.1155\,h^{-1}$
Original amount $A_{0}=0.050\,mg$
Amount remaining $A=6.3\times10^{-3}\,mg$
Recall that $\ln(\frac{A_{0}}{A})=kt$ where $t$ is the time required.
$\implies \ln(\frac{0.050\,mg}{6.3\times10^{-3}\,mg})=2.07147=0.1155\,h^{-1}(t)$
$\implies t=\frac{2.07147}{0.1155\,h^{-1}}=18\,h$