Answer
See the detailed answer below.
Work Step by Step
a) We know that the moment of inertia is given by
$$I=\int x^2 dm$$
And in our case, we are dealing with disks and radiuses, so
$$I=\int r^2 dm\tag 1$$
We know that the density of the disk is uniform, so
$$\rho=\dfrac{M}{V}=\dfrac{dm}{dV}$$
Hence,
$$ \dfrac{M}{ \color{red}{\bf\not}tA}=\dfrac{dm}{ \color{red}{\bf\not}tdA }$$
where $t$ is the thickness of the disk which is constant and $A$ is the area of the disk.
Thus,
$$dm=\dfrac{MdA}{ A}$$
Recall that the area of a disk's surface (the area of a circle) is given by $A=\pi r^2$. So that the area of the disk that is empty from the middle as our disk is given by $A=\pi (R^2-r^2)$.
Thus,
$$dm=\dfrac{MdA}{ \pi (R^2-r^2)} $$
where $dA$ is given by $2\pi r\times dr$
$$dm=\dfrac{M (2 \color{red}{\bf\not}\pi r)dr}{ \color{red}{\bf\not}\pi (R^2-r^2)} $$
Plugging into (1);
$$I=\int_r^R r^2 \dfrac{2M rdr}{ (R^2-r^2)} $$
$$I= \dfrac{M2 }{ (R^2-r^2)}\int_r^R r^3dr $$
Noting that $A_{disk}=A$ is constant and is given by $ \pi (R^2-r^2)$ that's why we pull it out of the integral.
$$I= \dfrac{2M }{ 4 (R^2-r^2)}\bigg|_r^R r^4 $$
$$I= \dfrac{M }{ 2(R^2-r^2)}(R^4-r^4)$$
$$I=\dfrac{M }{ 2 (R^2-r^2)}(R^2-r^2)(R^2+r^2)$$
$$\boxed{I=\dfrac{M(R^2+r^2) }{ 2}}\tag 2$$
_____________________________________________
b) When $r=0$,
$$I=\dfrac{M(R^2+0^2) }{ 2}=\boxed{\frac{1}{2}MR^2}$$
which is the same result of the moment of inertia of a sold disk as Table 12.2.
And when $r=R$,
$$I=\dfrac{M(R^2+R^2) }{ 2}=\boxed{ MR^2}$$
which is the same result of the moment of inertia of a cylindrical hoop, about center as Table 12.2.
_____________________________________________
c) We assume that the disk is rolling down the incline without friction which means that the energy is conserved.
Thus,
$$E_i=E_f$$
$$K_i+K_{i,rot}+U_{gi}=K_f+K_{f,rot}+U_{gf}$$
We choose at the bottom of the incline to be our origin at which $y=0$.
Thus, the final potential energy is zero.
$$K_i+K_{i,rot}+U_{gi}=K_f+K_{f,rot}+0$$
We know that the disk starts rolling from rest, so that the initial kinetic energy is zero.
$$0+0+U_{gi}=K_f+K_{f,rot}$$
$$ Mgh=\frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$$
Plugging $I$ from (2);
$$ \color{red}{\bf\not}Mgh=\frac{1}{2} \color{red}{\bf\not}Mv^2+\frac{1}{2}\left[\dfrac{ \color{red}{\bf\not}M(R^2+r^2) }{ 2}\right]\omega^2$$
$$ gh=\frac{1}{2} v^2+ \dfrac{ R^2+r^2 }{ 4} \omega^2$$
We can find the initial height $h$ from the right triangle below, $\sin\theta=\dfrac{h}{d}$. Hence, $h=d\sin\theta$;
$$ gd\sin\theta=\frac{1}{2} v^2+ \dfrac{ R^2+r^2 }{ 4} \omega^2$$
Recall that $v=\omega R$, so $\omega=v/R$.
Thus,
$$ gd\sin\theta=\frac{1}{2} v^2+ \dfrac{ R^2+r^2 }{ 4} \left(\dfrac{v}{R}\right)^2$$
$$ gd\sin\theta=\frac{1}{2} v^2+ \dfrac{ R^2+r^2 }{ 4} \left(\dfrac{v^2}{R^2}\right)$$
$$ gd\sin\theta=v^2\left[\frac{1}{2} + \dfrac{R^2+r^2 }{4R^2} \right]$$
Therfore, the velocity of the disk at the bottom of the incline is given by
$$v =\sqrt{\dfrac{gd\sin\theta}{\frac{1}{2} + \dfrac{R^2+r^2 }{4R^2} }}$$
Plugging the known;
$$v =\sqrt{\dfrac{9.8\times 0.5\sin20^\circ }{\frac{1}{2} + \dfrac{0.04^2+0.03^2 }{4\times 0.04^2} }}$$
$$v=\color{red}{\bf 1.372 }\;\rm m/s\tag 3$$
If the object was sliding rather than rolling, we can use the same conservation of fnenegy principle with no rotational kinetic energy.
Thus,
$$ \color{red}{\bf\not}Mgh=\frac{1}{2} \color{red}{\bf\not}M{v^2}'$$
$$v'=\sqrt{2gh}=\sqrt{2gd\sin\theta}$$
Plugging the known;
$$v'= \sqrt{2\times 9.8\times 0.5 \sin20^\circ}=\color{red}{\bf 1.831}\;\rm m/s$$
Therefore,
$$\dfrac{v}{v'}=\dfrac{1.372 }{1.831}=0.749\approx 0.75$$
$$v={\bf 0.75}v'$$