Answer
The wood cylinder wins the race.
Work Step by Step
We can write an expression for the rotational inertia of a solid sphere:
$I = \frac{1}{2}MR^2$
When the brass cylinder is hollowed out in the center, then we can express the rotational inertia as follows:
$I = cMR^2$, where $c \gt \frac{1}{2}$
We can use Equation (11-10) to find the acceleration of a solid cylinder down the incline:
$a = \frac{g~sin~\theta}{1+I/MR^2}$
$a = \frac{g~sin~\theta}{1+\frac{1}{2}MR^2/MR^2}$
$a = \frac{g~sin~\theta}{1+\frac{1}{2}}$
$a = \frac{2}{3}~g~sin~\theta$
We can use Equation (11-10) to find the acceleration of the hollowed out brass cylinder down the incline:
$a = \frac{g~sin~\theta}{1+I/MR^2}$
$a = \frac{g~sin~\theta}{1+c~MR^2/MR^2}$
$a = \frac{g~sin~\theta}{1+c} \lt \frac{g~sin~\theta}{1+\frac{1}{2}}=\frac{2}{3}~g~sin~\theta$
Since the acceleration of the brass cylinder is less than the acceleration of the wood cylinder, the wood cylinder reaches the bottom first.
The wood cylinder wins the race.
