To find the minimum number of attempts required to identify the lightest cube with certainty, we must consider the worst-case scenario using an optimal weighing strategy.

Step 1: First Weighing Divide the $7$ cubes into three groups: two groups of $3$ cubes each, and one group of $1$ cube. Place the two groups of $3$ cubes on either pan of the balance.

• Case 1: The balance is equal. This means all $6$ cubes on the balance are of equal weight. The lightest cube is the $1$ cube that was left out. Here, we find it in $1$ attempt.
• Case 2: The balance is unequal. One pan will be lighter, meaning the lightest cube is among the $3$ cubes on that lighter pan.

Step 2: Second Weighing Take the $3$ cubes from the lighter pan (from Case 2). Select any $2$ cubes and place one on each pan of the balance, leaving the $3$rd cube aside.

• Case 2a: The balance is equal. The lightest cube is the $1$ cube left aside.
• Case 2b: The balance is unequal. The cube on the lighter pan is the lightest cube.

In every possible scenario, a maximum of $2$ weighings is sufficient to guarantee finding the lightest cube. Therefore, the minimum number of attempts required to distinguish the odd cube with certainty is $2$.