Let the height of the sunflowers be a normally distributed random variable \(X\) with mean \(\mu = 115\) cm and standard deviation \(\sigma\).

We know that 80% of the heights are greater than 103 cm, which means 20% are less than 103 cm. This corresponds to a cumulative probability of 0.20.

Using the standard normal distribution table, we find that the z-score corresponding to a cumulative probability of 0.20 is approximately \(z = -0.842\).

We use the z-score formula:

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

Substituting the known values:

\(-0.842 = \frac{103 - 115}{\sigma}\)

\(-0.842 = \frac{-12}{\sigma}\)

Solving for \(\sigma\):

\(\sigma = \frac{-12}{-0.842}\)

\(\sigma \approx 14.3\)