Answer
$ -4\times10^{-3}$
Work Step by Step
We know that increasing the temperature increases the volume of the cylinder which means increases its radius. And increasing the radius means an increase in the moment of inertia of the cylinder which is given by
$I=\frac{1}{2}MR^2\tag 1$
Since the wheel is rotating around its own axis without friction, so there is no change in its angular momentum.
$$L_i=L_f$$
$$I_i\omega_i=I_f\omega_f$$
Solving for $\omega_f$ and using eq (1) above;
$$\omega_f=\dfrac{I_i\omega_i}{I_f}=\dfrac{\frac{1}{2}MR_i^2\omega_i}{\frac{1}{2}MR_f^2}$$
$$\omega_f=\dfrac{ R_i^2\omega_i}{ R_f^2}\tag 2$$
We know that the final radius is given by
$$R_f=R_i+R_i\alpha \Delta T=R_i\left(1+\alpha \Delta T\right)$$
Plugging into (2);
$$\omega_f=\dfrac{ R_i^2\omega_i}{ \left[R_i\left(1+\alpha \Delta T\right)\right]^2}=\dfrac{\omega_i}{\left(1+\alpha_{Al} \Delta T\right)^2}$$
Plugging the given and $\alpha_{AL}$ from table 13-1;
$$\omega_f=\dfrac{ 32.8}{\left(1+ [25\times10^{-6}\cdot 80]\right)^2}=\bf 32.67\;\rm rad/s$$
The fractional change in $\omega$;
$$\text{fractional change}=\dfrac{\Delta \omega}{\omega_i}=\dfrac{\omega_f-\omega_i}{\omega_i}=\dfrac{32.67-32.8}{32.8}$$
$$\text{fractional change}=-3.97\times10^{-3}\approx\bf -4\times10^{-3}$$
The negative sign is due to the decrease in the angular velocity due to the increase in radius length.