Correct : 5

Explanation:
To force the edge BC to appear in every minimum spanning tree (MST), there must be no alternative path from B to C whose largest edge weight is <= x.
If such a path exists, Kruskal's algorithm could connect B and C using that path's edges and skip the direct edge BC.
Examine all simple B–C paths and record the largest edge weight on each path:

• Path B – A – C: edge weights 7 and 1 → largest = 7
• Path B – D – C: edge weights 3 and 8 → largest = 8
• Path B – D – A – C: edge weights 3, 6, 1 → largest = 6

The minimum, over these paths, of the largest-edge value is 6. That means there exists a B–C path whose maximum edge is 6. For BC to be in every MST we need x < 6 (strictly less than 6), otherwise that path could be used instead of BC.
Since the answer asks for an integer value, the largest integer x satisfying x < 6 is: 5