Answer
The rotational inertia decreases by $~~64.3\%$
Work Step by Step
We can find an expression for the rotational inertia of the rod with three particles:
$I_0 = Md^2+M(2d)^2+M(3d)^2$
$I_0 = 14~Md^2$
We can find an expression for the rotational inertia of the rod with the outermost particle removed:
$I = Md^2+M(2d)^2$
$I = 5~Md^2$
We can find the ratio of $\frac{I}{I_0}$:
$\frac{I}{I_0} = \frac{5~Md^2}{14~Md^2} = 0.35714$
Note that $~~1.00 - 0.35714 = 0.643$
The rotational inertia decreases by $~~64.3\%$
