We start with the energy function:

$ E(t)= \alpha^3 e^{-\beta t}, $

where $\beta=0.3\, \text{s}^{-1}$.

Step 1. Take the Logarithm

Taking the logarithm of both sides:

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

Step 2. Differentiate to Find the Relative Error

Differentiate both sides:

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

For maximum error estimation, we consider the absolute values and sum the contributions:

$ \left|\frac{\Delta E}{E}\right| \approx 3\,\left|\frac{\Delta \alpha}{\alpha}\right| + \beta\, |\Delta t|. $

Step 3. Plug in the Given Errors

The percentage error in $\alpha$ is $1.2\%$ (i.e., $\Delta \alpha/\alpha = 0.012$).

The percentage error in $t$ is $1.6\%$, so for $t=5\,\text{s}$:

$ \Delta t = 0.016 \times 5 = 0.08\, \text{s}. $

Now substitute these values:

$ \left|\frac{\Delta E}{E}\right| \approx 3(0.012) + 0.3(0.08). $

Calculating each term:

$3(0.012) = 0.036$ (or $3.6\%$),

$0.3(0.08) = 0.024$ (or $2.4\%$).

Step 4. Compute the Total Maximum Percentage Error

Add the contributions:

$ \left|\frac{\Delta E}{E}\right| \approx 0.036 + 0.024 = 0.06, $

which is $6\%$.

Thus, the maximum percentage error in the energy at $t = 5\,\text{s}$ is:

$ \boxed{6\%} $

Answer: Option A (6%).