Step 1: Understanding the Concept:
For an ideal gas, the total translational kinetic energy is \(E = \frac32PV\). So the energy per unit volume is \(\frac{E}{V} = \frac32P\).
Step 2: Calculate:
Energy per litre: \(600\) J in \(1\) L \(= 10^{-3}\) m\(^3\), so \(\frac{E}{V} = 6\times10^5\) J/m\(^3\).
\[ P = \frac23\times\frac{E}{V} = \frac23\times6\times10^5 = 4\times10^5\ \text{N/m}^2 \]
Step 3: Other options:
\(6\times10^5\) is the energy density, not the pressure. \(3\times10^5\) would come from using \(P = \frac{E}{2V}\).
Final Answer:
The pressure is \(4\times10^5\) N/m\(^2\), option (A).
\[ \boxed{4\times10^5\ \text{N/m}^2} \]