B

Step 1: Identify forces acting on the system at the moment the thread is burnt. The gravitational force (weight) acting downwards on the rod is given by:
\[ W = mg = 2 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 19.62 \, \text{N} \]
Step 2: The spring force exerted at point B when the spring is extended by 2 cm (0.02 m) is given by Hooke's Law:
\[ F_s = kx = 600 \, \text{N/m} \times 0.02 \, \text{m} = 12 \, \text{N} \]
Step 3: The system is in static equilibrium just before the thread is burnt. The normal force (N) exerted by support O must balance the weight of the rod and the spring force:
\[ N + F_s = W \]
Rearranging gives us:
\[ N = W - F_s = 19.62 \, \text{N} - 12 \, \text{N} = 7.62 \, \text{N} \]
Therefore, the normal force exerted by the support at O is approximately 7.62 N, which rounds to 8 N (after properly considering response formats).
So, the closest correct answer is B.