Answer
$I = 8.52~kg~m/s^2$
Work Step by Step
Let $m_1$ be the mass of the disk and let $m_2$ be the mass of the outer ring.
We can find $m_1$.
$m_1 = (\pi~r_1^2)(3.00~g/cm^2)$
$m_1 = (\pi)(50.0~cm)^2(3.00~g/cm^2)$
$m_1 = 23.56~kg$
We can find $m_2$.
$m_2 = (\pi~r_2^2-\pi~r_1^2)(2.00~g/cm^2)$
$m_2 = [(\pi)(70.0~cm)^2-(\pi)(50.0~cm)^2](2.00~g/cm^2)$
$m_2 = 15.08~kg$
We can find the moment of inertia $I$
$I = \frac{1}{2}m_1~r_1^2+\frac{1}{2}m_2~(r_1^2+r_2^2)$
$I = \frac{1}{2}(23.56~kg)(0.500~m)^2+\frac{1}{2}(15.08~kg)[(0.500~m)^2+(0.700~m)^2]$
$I = 8.52~kg~m/s^2$
