Answer
$L=1.13Js$
Work Step by Step
As we know that
$I_{system}=2\times\frac{M_{rod}L^2}{12}+4\times M_{cup}(\frac{L}{2})^2$
We plug in the known values to obtain:
$I_{system}=2\times\frac{0.0757(0.326)^2}{12}+4\times(0.124)(\frac{0.326}{2})^2$
$I_{system}=14.5\times 10^{-3}Kgm^2$
We also know that
$L=I\omega$
$\implies L=14.5\times 10^{-3}\times 78=1.13Js$
