To find the amount of glucose required to prepare $250 \, \mathrm{mL}$ of a $\frac{M}{20}$ aqueous solution, we need to follow these steps:

1. Convert the volume from milliliters to liters, because molarity is expressed in moles per liter (M).

2. Use the equation for molarity:

$M = \frac{n}{V}$

where:

• $M$ is the molarity (given as $$\frac{1}{20} \, \mathrm{M}$$)
• $n$ is the number of moles of solute
• $V$ is the volume of the solution in liters

3. Rearrange the equation to solve for the number of moles of glucose ($n$):

$n = M \times V$

4. Convert the moles of glucose to grams using the molar mass of glucose.

Let's proceed with the calculations:

1. Convert the volume to liters:

$$250 \, \mathrm{mL} = \frac{250}{1000} \, \mathrm{L} = 0.25 \, \mathrm{L}$$

2. Calculate the moles of glucose:

$$n = \left( \frac{1}{20} \right) \times 0.25 \, \mathrm{L} = \frac{0.25}{20} \, \mathrm{mol} = 0.0125 \, \mathrm{mol}$$

3. Convert the moles of glucose to grams using the molar mass of glucose:

$$\text{mass} = n \times \text{molar mass}$$

$$\text{mass} = 0.0125 \, \mathrm{mol} \times 180 \, \mathrm{g} \, \mathrm{mol}^{-1} = 2.25 \, \mathrm{g}$$

So, the amount of glucose required is:

Option A: 2.25 g