Answer
$\rm 5.28\times 10^4\;pa$
Work Step by Step
To find the needed oil pressure, we need to find the force exerted on the wheel (or the disk) to bring it to a stop. And to find the force needed (the kinetic friction force), we need to find the angular acceleration of the disk that brings it to a stop in 5 seconds.
$\bullet$ The angular acceleration of the disk is given by
$$\alpha =\dfrac{\omega_f-\omega_i}{t}$$
Plugging the given and remember to convert the initial angular speed of the wheel to rad/s from rpm.
$$\alpha =\dfrac{0-\rm \left[\dfrac{900\;\rm rev}{1\;min}\right]\left[\dfrac{1\;min}{60\;s}\right]\left[\dfrac{2\pi\;rad}{1\;rev}\right]}{5\;\rm s}$$
$$\alpha= {\bf-6\pi}\;\rm rad/s\tag 1$$
$\bullet$ The force needed is given by
$$\sum\tau_{disk}=-2f_kd=I\alpha$$
The 2 in front of the force of friction is due to the two forces exerted by the two brake pads. $I$ is the moment of inertia of a disk $\frac{1}{2}MR^2$ and $f_k=\mu_kF_n$ where $F_n$ is the normal force needed. $d$ is the distance from the axis of the disk to the pads.
$$-2\mu_kF_nd=\left(\frac{1}{2}MR^2\right) \alpha$$
Plugging from (1);
$$-2\mu_kF_nd=\left(\frac{1}{2}MR^2\right) (-6\pi) $$
Hence,
$$F_n=\dfrac{3\pi MR^2}{2\mu_k d}\tag 2$$
$\bullet$ The oil pressure needed is given by
$$P=\dfrac{F_n}{A}$$
where $A$ is the cross-sectional area of the brake pad.
Plugging from (2);
$$P=\dfrac{\dfrac{3\pi MR^2}{2\mu_k d}}{\pi r^2}$$
$$P= \dfrac{3\color{red}{\bf\not}\pi MR^2}{2\mu_k\color{red}{\bf\not} \pi r^2d} $$
$$P= \dfrac{3 MR^2}{2\mu_k r^2d} $$
Plugging the known;
$$P= \dfrac{3 (15)(0.13)^2}{2(0.6)(0.01)^2(0.12)} $$
$$P=\color{red}{\bf 5.28\times 10^4}\;\rm Pa$$
