Step-by-Step Solution:

1. Write the balanced chemical equation for the electrolysis of water. The electrolysis of liquid water ($H_2O(l)$) produces hydrogen gas ($H_2(g)$) and oxygen gas ($O_2(g)$). The balanced chemical equation for the decomposition of one mole of water is: $H_2O(l) \rightarrow H_2(g) + \frac{1}{2}O_2(g)$

2. Calculate the number of moles of water being electrolyzed. First, we determine the molar mass of water ($H_2O$). Molar mass of H = 1 g/mol Molar mass of O = 16 g/mol Molar mass of $H_2O = 2 \times (1) + 16 = 18$ g/mol.

   Given mass of water = 144 g. The number of moles of water ($n_{H_2O}$) is: $n_{H_2O} = \frac{\text{mass}}{\text{molar mass}} = \frac{144 \text{ g}}{18 \text{ g/mol}} = 8 \text{ mol}$

3. Determine the change in the number of moles of gas ($\Delta n_g$). The overall reaction for the electrolysis of 8 moles of water is: $8 H_2O(l) \rightarrow 8 H_2(g) + 4 O_2(g)$ The initial state has 8 moles of liquid water, so the initial number of moles of gas is $n_{g, initial} = 0$. The final state consists of gaseous products. The total number of moles of gas in the final state is: $n_{g, final} = n_{H_2} + n_{O_2} = 8 \text{ mol} + 4 \text{ mol} = 12 \text{ mol}$ The change in the number of moles of gas is: $\Delta n_g = n_{g, final} - n_{g, initial} = 12 - 0 = 12 \text{ mol}$

4. Calculate the expansion work done. The work done on the system during a process at constant pressure and temperature is given by the formula: $w = -P_{ext}\Delta V$ Assuming the gases behave ideally and the external pressure is equal to the gas pressure, we can use the ideal gas law, $PV = nRT$. For a change in the number of moles of gas at constant T and P, this becomes: $P\Delta V = (\Delta n_g)RT$ Substituting this into the work equation, we get: $w = -\Delta n_g RT$

5. Substitute the values and compute the work. We are given: $\Delta n_g = 12$ mol $R = 8.3$ J K⁻¹ mol⁻¹ $T = 300$ K

   Now, we calculate the work in Joules: $w = -(12 \text{ mol}) \times (8.3 \text{ J K⁻¹ mol⁻¹}) \times (300 \text{ K})$ $w = -12 \times 8.3 \times 300 \text{ J}$ $w = -12 \times 2490 \text{ J}$ $w = -29880 \text{ J}$

6. Convert the work from Joules (J) to kiloJoules (kJ). To convert Joules to kiloJoules, we divide by 1000: $w = \frac{-29880}{1000} \text{ kJ} = -29.88 \text{ kJ}$

   The negative sign indicates that work is done by the system on the surroundings (expansion). The question asks for the "expansion work done". This can be interpreted either as the work done on the system ($w = -29.88$ kJ) or the work done by the system ($w_{by} = -w = 29.88$ kJ). The provided answer range accepts both positive and negative values. Our calculated value of 29.88 falls within the range [29.8, 29.95] and -29.88 falls within [-29.95, -29.8].

Final Answer: The expansion work done is -29.88 kJ.