Initially in the container, only the reactant $\mathrm{A}$ is present. After $5 \text{hours}$ equilibrium is established.

$$\begin{gathered} \mathrm{A} \rightleftharpoons \mathrm{nB} \\ \mathrm{Initially} (\mathrm{t}=0) 0.6 \mathrm{M} 0 \\ \mathrm{At} \mathrm{equilibrium} (\mathrm{t}=5 \mathrm{hr}) 0.3 \mathrm{M} 0.6 \mathrm{M} \end{gathered}$$

At equilibrium, the rate of forward reaction becomes equal to the rate of backward reaction and the net rate of the reaction is zero.

$$\begin{gathered} -\frac{{\displaystyle ∆\left[\mathrm{A}\right]}}{{\displaystyle ∆\mathrm{t}}}=+\frac{1}{\mathrm{n}}\times \frac{∆\left[\mathrm{B}\right]}{∆\mathrm{t}} \\ -\frac{{\displaystyle \left[0.3-0.6\right]}}{{\displaystyle \left[5-0\right]}}=+\frac{1}{\mathrm{n}}\times \frac{\left[0.6-0\right]}{\left[5-0\right]} \\ \mathrm{n}=2 \end{gathered}$$