The problem involves determining the composition of an octasaccharide that, upon complete hydrolysis, yields three types of monosaccharides: ribose, 2-deoxyribose, and glucose.

First, consider the balanced chemical equation for the hydrolysis of the octasaccharide:

$ \text{Octasaccharide} + 7 \text{H}_2\text{O} \rightarrow \text{Ribose} + \text{2-deoxyribose} + \text{Glucose} $

The initial molar mass of the octasaccharide is $1024 \, \text{g/mol}$, and it requires 7 water molecules (each with a molar mass of $18 \, \text{g/mol}$, thus totaling $126 \, \text{g/mol}$) to undergo hydrolysis. Therefore, the total mass on the reactant side is:

$ 1024 + 126 = 1150 \, \text{g} $

According to the given data, the 2-deoxyribose formed constitutes $58.26\%$ of the total mass of the monosaccharides. To find the mass of 2-deoxyribose, calculate:

$ 1150 \times \frac{58.26}{100} = 669.99 \, \text{g} \approx 670 \, \text{g} $

The molar mass of 2-deoxyribose is $134 \, \text{g/mol}$, so the number of units produced is:

$ \frac{670}{134} = 5 \, \text{units} $

Assuming there is one unit of glucose (molar mass $180 \, \text{g/mol}$), the remaining units in the octasaccharide must be ribose. Given five units of 2-deoxyribose and one unit of glucose, the potential number of ribose units can be calculated as the difference to reach a total of eight saccharide units, ensuring:

$ 5 \text{ (2-deoxyribose)} + 1 \text{ (glucose)} + x \text{ (ribose)} = 8 $

Solving this gives:

$ x = 8 - 5 - 1 = 2 $

To verify the setup, the total mass of the hydrolysis products equals the mass at the reactant side:

$ 670 \, \text{(2-deoxyribose)} + 180 \, \text{(glucose)} + (2 \times 150 \, \text{(ribose)}) = 1150 \, \text{g} $

Thus, the octasaccharide contains 2 ribose units. Therefore, the number of ribose units present in one molecule of the octasaccharide is 2.