Answer
a) $\sum I=7182kg.m^2$
b) $KE_R=6.95\times10^6J$
Work Step by Step
a) The blade is a thin rod having the rotation axis at one end, so $$I_{blade}=\frac{1}{3}MR^2=\frac{1}{3}(240kg)(6.7m)^2=3591kg.m^2$$
The total moment of inertia of 2 blades is $\sum I=2I_{blade}=7182kg.m^2$
b) The rotational kinetic energy of the spinning blades is
$$KE_R=\frac{1}{2}\sum I\omega^2$$
We have $\omega=44rad/s$, so $$KE_R=6.95\times10^6J$$
