Step 1: Concept
The work done during the expansion of a gas against a constant external pressure is given by the formula $W = -P_{ext} \Delta V$.
Step 2: Meaning
Work is expressed in Joules ($J$), and pressure in bars. Note that $1 \text{ bar dm}^3 = 100 \text{ J}$. The change in volume $\Delta V$ is $V_2 - V_1$.
Step 3: Analysis
Given: $W = -600 \text{ J}$, $V_1 = 15 \text{ dm}^3$, $V_2 = 20 \text{ dm}^3$.
$\Delta V = 20 - 15 = 5 \text{ dm}^3$.
$W \text{ in bar dm}^3 = -600 / 100 = -6 \text{ bar dm}^3$.
$-6 = -P_{ext} \times 5$.
$P_{ext} = 6 / 5 = 1.2 \text{ bar}$.
Step 4: Conclusion
The constant external pressure required is $1.2 \text{ bar}$.
Final Answer: (A)