Answer
$I = 0.008352~kg~m^2$
Work Step by Step
We can use the parallel axis theorem to find the rotational inertia:
$I = (\frac{1}{2})(\frac{M}{15})(\frac{L}{30})^2+(2)\times[(\frac{1}{2})(\frac{M}{15})(\frac{L}{30})^2+(\frac{M}{15})(\frac{L}{15})^2]+...+(2)\times [(\frac{1}{2})(\frac{M}{15})(\frac{L}{30})^2+(\frac{M}{15})(\frac{7L}{15})^2]$
$I = (\frac{15}{2})(\frac{M}{15})(\frac{L}{30})^2+(2)(\frac{M}{15})(\frac{L}{15})^2+...+(2)(\frac{M}{15})(\frac{7L}{15})^2$
$I = (\frac{15}{2})(\frac{M}{15})(\frac{L}{30})^2+(2)(\frac{M}{15})(\frac{L}{15})^2~(1^2+2^2+...+7^2)$
$I = (\frac{15}{2})(\frac{M}{15})(\frac{L}{30})^2+(280)(\frac{M}{15})(\frac{L}{15})^2$
$I = (\frac{15}{2})(\frac{0.1000~kg}{15})(\frac{1.0000~m}{30})^2+(280)(\frac{0.1000~kg}{15})(\frac{1.0000~m}{15})^2$
$I = 0.008352~kg~m^2$
