We are given that the weight of the tea packets follows a normal distribution with mean \(\mu = 260\) g and standard deviation \(\sigma\) g. A packet is considered underweight if it weighs less than 250 g, and 1% of packets are underweight.

We need to find the value of \(\sigma\) such that the probability of a packet being underweight is 0.01. This corresponds to the left tail of the normal distribution.

Using the standard normal distribution table, the z-score corresponding to the 1st percentile is approximately \(z = -2.326\).

We use the standardization formula:

\(z = \frac{X - \mu}{\sigma}\)

Substituting the known values:

\(-2.326 = \frac{250 - 260}{\sigma}\)

\(-2.326 = \frac{-10}{\sigma}\)

Solving for \(\sigma\):

\(\sigma = \frac{-10}{-2.326}\)

\(\sigma \approx 4.30\)