To find the mass of sodium hydroxide (NaOH) left unreacted, we first need to determine the moles of NaOH and HCl originally present and compare them to see which one is in excess.

The molar mass of NaOH is approximately $40 \text{ g/mol}$, so the number of moles of NaOH in 1 gram can be calculated as follows:

$$\text{Moles of NaOH} = \frac{\text{Mass of NaOH}}{\text{Molar mass of NaOH}} = \frac{1 \text{ g}}{40 \text{ g/mol}} = 0.025 \text{ moles}$$

Next, we calculate the number of moles of HCl using its concentration and the volume of the solution. Recall that concentration (Molarity, M) is defined as moles of solute per liter of solution. Given that the concentration of HCl is $0.75 \text{ M}$ and the volume of the solution is $25 \text{ mL}$ or $0.025 \text{ L}$, we can find the moles of HCl:

$$\text{Moles of HCl} = \text{Concentration} \times \text{Volume in liters} = 0.75 \text{ M} \times 0.025 \text{ L} = 0.01875 \text{ moles}$$

Now, we compare the moles of NaOH and HCl. The stoichiometry of the reaction between NaOH and HCl is:

$$\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$$

Each mole of NaOH reacts with one mole of HCl. Given that there are more moles of NaOH (0.025 moles) than HCl (0.01875 moles), HCl is the limiting reactant and will be completely consumed. The excess NaOH can be calculated:

$$\text{Excess moles of NaOH} = \text{Moles of NaOH} - \text{Moles of HCl} = 0.025 \text{ moles} - 0.01875 \text{ moles} = 0.00625 \text{ moles}$$

To find the mass of the unreacted NaOH:

$$\text{Mass of unreacted NaOH} = \text{Moles of unreacted NaOH} \times \text{Molar mass of NaOH} = 0.00625 \text{ moles} \times 40 \text{ g/mol} = 0.25 \text{ g} = 250 \text{ mg}$$

Therefore, the mass of sodium hydroxide left unreacted is 250 mg, which corresponds to Option B.