Given that 96% of ferry crossings take longer than \(t\), this implies that 4% take less time than \(t\). We need to find the corresponding \(z\)-value for the 4th percentile of the standard normal distribution.

The \(z\)-value for 4% in the standard normal distribution is approximately \(z = -1.751\).

Using the standardization formula:

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

Substitute \(z = -1.751\), \(\mu = 85\), and \(\sigma = 6.8\):

\(-1.751 = \frac{t - 85}{6.8}\)

Solving for \(t\):

\(t - 85 = -1.751 \times 6.8\)

\(t - 85 = -11.9088\)

\(t = 85 - 11.9088\)

\(t = 73.0912\)

Rounding to one decimal place, \(t = 73.1\).