The given bar graph represents the utilization of ATP and NADPH...

The Calvin cycle consists of three main stages: Carboxylation, Reduction, and Regeneration.

1. Carboxylation (Stage I): In this step, $CO_{2}$ combines with Ribulose-1,5-bisphosphate (RuBP) to form two molecules of 3-phosphoglycerate (3-PGA). This step does not utilize any ATP or NADPH. This matches the graph for Stage I, where the utilization is zero.

2. Reduction (Stage II): The two molecules of 3-PGA are reduced to form triose phosphate. This process requires 2 molecules of ATP (for phosphorylation) and 2 molecules of NADPH (for reduction) per molecule of $CO_{2}$ fixed. This perfectly matches the graph for Stage II, which shows 2 ATP and 2 NADPH utilized.

3. Regeneration (Stage III): The $CO_{2}$ acceptor molecule, RuBP, is regenerated from triose phosphate to keep the cycle going. This step requires 1 molecule of ATP for phosphorylation per molecule of $CO_{2}$ fixed, and no NADPH is used. This matches the graph for Stage III, which shows 1 ATP and 0 NADPH utilized.

Therefore, the stages are I - Carboxylation, II - Reduction, and III - Regeneration.

Answer: I - Carboxylation, II - Reduction, III - Regeneration