Answer
13.2 hours
Work Step by Step
Amount of radionuclide at the beginning $A_{0}=0.4\,mg$
Amount of radionuclide remaining
$A=0.1\,mg$
Time $t=26.4\,h$
$\ln(\frac{A_{0}}{A})=kt=\frac{0.693}{t_{1/2}}\times t$
$\implies \ln(\frac{0.4}{0.1})=1.386=\frac{0.693}{t_{1/2}}\times 26.4\,h$
Then, $t_{1/2}=\frac{0.693\times26.4\,h}{1.386}=13.2\,h$
