Solution:
$\text{Hint: All units of volume are applicable}$
$\text{Step 1: Consider the formula for work in reversible isothermal expansion}$
$\text{The work done during reversible isothermal expansion of an ideal gas is given by:}$
$W_{\text{rev}} = -nRT\ln\frac{V_2}{V_1}$
$\text{or using base-10 logarithm:}$
$W_{\text{rev}} = -2.303\,nRT\log\frac{V_2}{V_1}$
$\text{Step 2: Analyze the units in this formula}$
$\text{In this formula:}$
$\bullet \, n \text{ is the number of moles (mol)}$
$\bullet \, R \text{ is the universal gas constant (J mol}^{-1}\text{ K}^{-1}\text{)}$
$\bullet \, T \text{ is the temperature (K)}$
$\bullet \, \frac{V_2}{V_1} \text{ is the ratio of final volume to initial volume}$
$\text{Step 3: Examine the volume ratio term}$
$\text{Looking at the volume ratio } \frac{V_2}{V_1}\text{, notice that:}$
$\text{• This is a dimensionless ratio}$
$\text{• The units of } V_2 \text{ and } V_1 \text{ must be the same, but the specific unit doesn't matter}$
$\text{• Whether volumes are expressed in m}^3\text{, dm}^3\text{, or cm}^3\text{, the ratio remains the same}$
$\text{For example, if } V_1 = 1\text{ m}^3 \text{ and } V_2 = 2\text{ m}^3\text{, the ratio is 2.}$
$\text{Similarly, if } V_1 = 1000\text{ dm}^3 \text{ and } V_2 = 2000\text{ dm}^3\text{, the ratio is still 2.}$
$\text{And if } V_1 = 1,000,000\text{ cm}^3 \text{ and } V_2 = 2,000,000\text{ cm}^3\text{, the ratio remains 2.}$
$\text{Step 4: Confirm the units of the final result}$
$\text{The units of the terms in the work equation are:}$
$\bullet \, n \text{ in mol}$
$\bullet \, R \text{ in J mol}^{-1}\text{ K}^{-1}$
$\bullet \, T \text{ in K}$
$\bullet \, \ln\frac{V_2}{V_1} \text{ is dimensionless}$
$\text{Therefore, the work will be in joules regardless of which volume unit is used, as long as the same unit is used for both } V_1 \text{ and } V_2\text{.}$
$\text{Hence, the volume can be expressed in m}^3\text{, dm}^3\text{, cm}^3\text{, or any other unit of volume, as long as the same unit is used consistently.}$
$\text{The correct answer is: All of the above.}$