To determine the amount of oxygen required for the complete combustion of 900 g of glucose, we need to follow these steps:

First, let's write the balanced chemical equation for the combustion of glucose $$(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6)$$:

$$\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 + 6 \mathrm{O}_2 \rightarrow 6 \mathrm{CO}_2 + 6 \mathrm{H}_2 \mathrm{O}$$

From the balanced equation, we see that 1 mole of glucose reacts with 6 moles of oxygen $$(\mathrm{O}_2)$$.

Next, let's calculate the number of moles of glucose in 900 g. The molar mass of glucose is given as $$180 \ \mathrm{g} \ \mathrm{mol}^{-1}$$.

Number of moles of glucose = $$\frac{900 \ \mathrm{g}}{180 \ \mathrm{g} \ \mathrm{mol}^{-1}} = 5 \ \mathrm{mol}$$

According to the balanced equation, 1 mole of glucose requires 6 moles of oxygen. Therefore, 5 moles of glucose will require:

Number of moles of oxygen = $$5 \ \mathrm{mol} \times 6 = 30 \ \mathrm{mol}$$

Now, we need to find the mass of 30 moles of oxygen. The molar mass of oxygen $$\left(\mathrm{O}_2\right)$$ is $$32 \ \mathrm{g} \ \mathrm{mol}^{-1}$$.

Mass of oxygen = $$30 \ \mathrm{mol} \times 32 \ \mathrm{g} \ \mathrm{mol}^{-1} = 960 \ \mathrm{g}$$

Therefore, the amount of oxygen required for the complete combustion of 900 g of glucose is 960 g.

So, the correct option is Option D: 960.