The correct answer is $\boxed{\text{B) 150 J}}$.
The kinetic energy of a rolling object is given by the equation:
$$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$
where $m$ is the mass of the object, $v$ is the velocity of the center of mass, $I$ is the moment of inertia, and $\omega$ is the angular velocity.
In this case, we are given that $m = 4\text{ kg}$, $v = 5\text{ m/s}$, and $I = 3\text{ kg m}^2$. We can calculate the angular velocity from the equation:
$$\omega = \frac{v}{r}$$
where $r$ is the radius of the object. In this case, $r = 0.5\text{ m}$, so $\omega = 10\text{ rad/s}$.
Substituting these values into the equation for kinetic energy, we get:
$$KE = \frac{1}{2}(4\text{ kg})(5\text{ m/s})^2 + \frac{1}{2}(3\text{ kg m}^2)(10\text{ rad/s})^2 = 150\text{ J}$$
Therefore, the kinetic energy of the disc is $\boxed{\text{150 J}}$.
Option A is incorrect because it is the kinetic energy of a point mass with the same mass and velocity as the disc. Option C is incorrect because it is the kinetic energy of a solid cylinder with the same mass and radius as the disc. Option D is incorrect because it is the kinetic energy of a hollow cylinder with the same mass and radius as the disc.