Answer
The magnitude of the normal force is $0.78N$
Work Step by Step
The front wheel has an initial angular velocity $\omega_0=13.1rad/s$ and final velocity $\omega=3.7rad/s$. The duration the brake is applied is $t=3s$. The angular acceleration is $$\alpha=\frac{\omega-\omega_0}{t}=-3.13rad/s^2$$
The net torque applied to the rim to have that deceleration is $$\sum\tau=(\sum mr^2)\alpha$$
Here, all the mass is concentrated on the rim, so $\sum mr^2=(1.3kg)(0.33m)^2=0.142$
$$\sum\tau=-0.44N.m$$
This net torque is produced by kinetic friction on 2 brake pads on 2 sides. So the torque on 1 break pad is $\tau=-0.22N.m$
$f_k$ produces this torque with a lever arm equal to the wheel's radius, so $$\tau=f_kr=\mu_kNr$$ $$N=\frac{\tau}{\mu_kr}=-0.78N$$
The magnitude of the normal force is $0.78N$
