The tension in the cord connecting B and C when the assembly was released is the weight of C. To find the weight, we must find the acceleration of the system.
Direct x-axis to the right, y-axis upward. Write Newton's second law for box A:
We can consider the hanging boxes B and C as one body since the cord is non-inextensible. Therefore, Newton's second law for them looks like
Therefore, the tension between B and C is
To answer how far A goes in the first 0.25 s, all we need is to use this simple equation: