Answer
$$ 0.205 \mathrm{kg} \cdot \mathrm{m}^{2}$$
Work Step by Step
The total rotational inertia of the disk and rod is
$\begin{aligned} I &=\frac{1}{2} M r^{2}+M(L+r)^{2}+\frac{1}{3} m L^{2} \\ &=\frac{1}{2}(0.500 \mathrm{kg})(0.100 \mathrm{m})^{2}+(0.500 \mathrm{kg})(0.500 \mathrm{m}+0.100 \mathrm{m})^{2}+\frac{1}{3}(0.270 \mathrm{kg})(0.500 \mathrm{m})^{2} \\&= 0.205 \mathrm{kg} \cdot \mathrm{m}^{2} \end{aligned}$
