Answer
The angular speed of the resultant combination is $~~267~rev/min$
Work Step by Step
Let $I_w$ be the rotational inertia of the first wheel.
Then $2I_w$ is the rotational inertia of the second wheel.
We can use conservation of angular momentum to find the angular speed of the resultant combination:
$L_f = L_i$
$(I_w+2I_w)~\omega_f = I_w~\omega_i$
$3I_w~\omega_f = I_w~\omega_i$
$\omega_f =\frac{\omega_i}{3}$
$\omega_f =\frac{800~rev/min}{3}$
$\omega_f = 267~rev/min$
The angular speed of the resultant combination is $~~267~rev/min$
