For the complete disk ...
mass = (area)(area density)
$m_1 = \pi(50 \, cm)^2(3 \, g/cm^2) = 7500\pi \, g = 7.5\pi \, kg$
for the concentric ring, I used a subtraction technique to calculate its rotational inertia.
$I_{ring}$ = ($I$ of a whole disk if the "hole" were filled)  ($I$ of the smaller disk "hole")
mass of the "whole" disk = (area)(area density)
$M = \pi(70 \, cm)^2(2 \, g/cm^2) = 9800\pi \, g = 9.8\pi \, kg$
mass of the "hole" disk = (area)(area density)
$m = \pi(50 \, cm)^2(2 \, g/cm^2) = 5000\pi \, g = 5\pi \, kg$
$I_{ring} = \dfrac{1}{2}MR^2  \dfrac{1}{2}mr^2$
