Answer
The wheel's angular speed is $~~0.17~rad/s$
Work Step by Step
We can find the angular velocity of the train with respect to the track:
$\omega = \frac{v}{R} = \frac{0.15~m/s}{0.43~m} = 0.3488~rad/s$
Let $\omega_w$ be the angular velocity of the wheel. Then we can let the angular velocity of the train be $~~\omega_t = \omega_w-0.3488~rad/s$
We can use conservation of angular momentum to find the angular velocity of the track:
$L_f = L_i$
$I_w~\omega_w+I_t~\omega_t = 0$
$(1.1m)R^2~\omega_w+(m)R^2~\omega_t = 0$
$1.1~\omega_w+(\omega_w-0.3488~rad/s) = 0$
$2.1~\omega_w = 0.3488~rad/s$
$\omega_w = \frac{0.3488~rad/s}{2.1}$
$\omega_w = 0.17~rad/s$
The wheel's angular speed is $~~0.17~rad/s$
