Let the mean be \(\mu = 9.3\) and the standard deviation be \(\sigma\). We are given that \(P(X > 5.6) = 0.85\).

Using the standard normal distribution, we find \(P(Z > z) = 0.85\), which implies \(P(Z < z) = 0.15\).

From the standard normal distribution table, \(z \approx -1.036\).

Using the formula for the standard normal variable, \(z = \frac{X - \mu}{\sigma}\), we have:

\(z = \frac{5.6 - 9.3}{\sigma} = -1.036\)

Solving for \(\sigma\):

\(-1.036 = \frac{5.6 - 9.3}{\sigma}\)

\(\sigma = \frac{5.6 - 9.3}{-1.036}\)

\(\sigma = 3.57\)