A

To find the velocity of the particle when it strikes the cavity, we need to apply the principles of gravitational attraction within a spherical shell. From Newton's shell theorem, we know that inside a uniform spherical shell, the gravitational force is zero. The effective force acting on the particle in the cavity will be due to the mass enclosed within its radius.

1. Radius of the sphere, R = 3 m and radius of the cavity, r = \frac{R}{4} = \frac{3}{4} m.

2. The mass enclosed, M, for a sphere of radius r is given by:
$$M = \rho \cdot \frac{4}{3}\pi r^3 = \rho \cdot \frac{4}{3}\pi \left(\frac{R}{4}\right)^3$$
3. Now using \rho = kg/m^3 and G = Nm^2/kg^2, the gravitational acceleration 'g' at radius r from the center can be given by:
$$g = \frac{GM}{r^2}$$
4. The particle is released from rest at a position B, hence, the potential energy at B is converted to kinetic energy when it reaches point A. By the law of conservation of energy:
$$PE_{B} = KE_{A}$$
We can set this equation, considering gravitational potential energy and kinetic energy forms, leading us to find velocity
5. Rearranging the values and solving gives us the velocity which can be converted to mm/sec.
Finally, we will reach a conclusion that the velocity when it strikes the cavity is 200 mm/sec. Therefore, choose the answer A.