Solution

Using Kruskal's algorithm, edges are selected in increasing order of weight.

From the graph:

$a-b = 1$

$d-f = 1$

$b-f = 2$

$d-c = 2$

$d-e = 3$

So, the valid order must follow:

Weight $1$ edges first:

$I, II$ in any order

Then weight $2$ edges:

$III, IV$ in any order

Then weight $3$ edge:

$V$

Now check option (c):

II, I, III, V, IV

Here, edge $V = d-e$ has weight $3$, but edge $IV = d-c$ has weight $2$.

So, selecting $V$ before $IV$ violates Kruskal's increasing weight rule.

Therefore, this order cannot be used.