Let the random variable representing the time spent be denoted by \(X\), where \(X \sim N(96, 18^2)\).

We need to find \(P(85 < X < 100)\).

Standardize the variable: \(Z = \frac{X - 96}{18}\).

Thus, \(P(85 < X < 100) = P\left(\frac{85 - 96}{18} < Z < \frac{100 - 96}{18}\right)\).

Calculate the standardized values: \(\frac{85 - 96}{18} = -0.6111\) and \(\frac{100 - 96}{18} = 0.2222\).

Therefore, \(P(-0.6111 < Z < 0.2222) = \Phi(0.2222) + \Phi(0.6111) - 1\).

Using standard normal distribution tables, \(\Phi(0.2222) = 0.5879\) and \(\Phi(0.6111) = 0.7294\).

Thus, \(P(-0.6111 < Z < 0.2222) = 0.5879 + 0.7294 - 1 = 0.3173\).

Rounding to three decimal places, the probability is \(0.317\).