Answer
$I = 0.023~kg~m^2$
Work Step by Step
We can consider the two rods as a single rod with a length $2d$
We can find the rotational inertia:
$I = md^2+m(2d)^2+\frac{1}{3}(2M)(2d)^2$
$I = 5md^2+\frac{8}{3}Md^2$
$I = (5)(0.85~kg)(0.056~m)^2+\frac{8}{3}(1.2~kg)(0.056~m)^2$
$I = 0.023~kg~m^2$
