Answer
$\frac{2}{3}~~$ of the original rotational kinetic energy is lost.
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 an expression for the angular speed of the resultant combination:
$L_f = L_i$
$(I_w+2I_w)~\omega_f = I_w~\omega_i$
$3I_w~\omega_f = I_w~\omega_i$
$\omega_f =\frac{\omega_i}{3}$
We can write an expression for the initial rotational kinetic energy:
$K_i = \frac{1}{2}I_w\omega_i^2$
We can write an expression for the new rotational kinetic energy:
$K_f = \frac{1}{2}I_f\omega_f^2$
$K_f = \frac{1}{2}(3I_w)(\frac{\omega_i}{3})^2$
$K_f = \frac{1}{3}\times \frac{1}{2}I_w\omega_i^2$
$K_f = \frac{1}{3}~K_i$
Therefore, $~~\frac{2}{3}~~$ of the original rotational kinetic energy is lost.
