Let the random variable \(X\) represent the weights of the bags of sugar. We know \(X\) is normally distributed with mean \(\mu = 1.04\) kg and standard deviation \(\sigma = 0.06\) kg.

We are given that \(P(X < w) = 0.81\). To find \(w\), we standardize \(X\) to a standard normal variable \(Z\):

\(P\left(Z < \frac{w - 1.04}{0.06}\right) = 0.81\)

Using the standard normal distribution table, we find that the z-score corresponding to 0.81 is approximately 0.878.

Thus, we have:

\(\frac{w - 1.04}{0.06} = 0.878\)

Solving for \(w\):

\(w - 1.04 = 0.878 \times 0.06\)

\(w - 1.04 = 0.05268\)

\(w = 1.04 + 0.05268\)

\(w = 1.09268\)

Therefore, \(1.09 \leq w \leq 1.093\).