The correct answer is $\boxed{\text{B) 0.05 m/sec}}$.
According to the law of conservation of momentum, the total momentum of the system before the bullet is fired is equal to the total momentum of the system after the bullet is fired. In this case, the system consists of the bullet and the gun. The initial momentum of the system is zero, since the bullet and the gun are at rest. The final momentum of the system is the momentum of the bullet plus the momentum of the gun. The momentum of the bullet is $p_b = mv_b = 0.2 \text{ kg} \times 25 \text{ m/s} = 5 \text{ kg m/s}$. The momentum of the gun is $p_g = mv_g$, where $v_g$ is the velocity of recoil of the gun. We can write the law of conservation of momentum as follows:
$$p_b + p_g = 0$$
$$5 \text{ kg m/s} + p_g = 0$$
$$p_g = -5 \text{ kg m/s}$$
The negative sign indicates that the velocity of recoil of the gun is in the opposite direction of the velocity of the bullet. The velocity of recoil of the gun is therefore $v_g = \frac{p_g}{m_g} = \frac{-5 \text{ kg m/s}}{100 \text{ kg}} = -0.05 \text{ m/s}$.
The other options are incorrect because they do not give the correct value for the velocity of recoil of the gun.