1. Given expression

   $E(t)=\alpha^3 e^{-\beta t}, \qquad \beta=0.3\,\text{s}^{-1}$

   We need the maximum percentage error in $E$ at $t=5\,$s.

2. Use logarithmic differentiation for error propagation

   Taking logarithm:

   $\ln E = 3\ln \alpha - \beta t$

   Differentiating:

   $\frac{dE}{E} = 3\frac{d\alpha}{\alpha} - \beta \, dt$

   For maximum error, we take magnitudes and add:

   $\left|\frac{\Delta E}{E}\right|_{\max} = 3\left|\frac{\Delta \alpha}{\alpha}\right| + \beta \, \Delta t$

3. Error due to $\alpha$

   Given percentage error in $\alpha$ is $1.2\%$.

   Since $E \propto \alpha^3$,

   $\text{percentage error due to } \alpha = 3\times 1.2\% = 3.6\%$

4. Error due to $t$

   Given percentage error in $t$ is $1.6\%$ at $t=5\,$s.

   So absolute error in time:

   $\Delta t = 1.6\% \text{ of } 5 = \frac{1.6}{100}\times 5 = 0.08\,\text{s}$

   Now,

   $\beta \Delta t = 0.3\times 0.08 = 0.024$

   Converting to percentage:

   $0.024\times 100 = 2.4\%$

5. Total maximum percentage error

   $\text{Maximum percentage error in } E = 3.6\% + 2.4\% = 6.0\%$

6. Check options

   • A: $6\%$ ✅
   • B: $11.6\%$ ❌
   • C: $4\%$ ❌
   • D: $8.4\%$ ❌

Therefore, the correct answer is A.