We need to find \(P(1.98 < X < 2.03)\) where \(X \sim N(2.02, 0.03^2)\).

First, standardize the values using the formula \(z = \frac{x - \mu}{\sigma}\).

For \(x = 1.98\), \(z = \frac{1.98 - 2.02}{0.03} = -1.333\).

For \(x = 2.03\), \(z = \frac{2.03 - 2.02}{0.03} = 0.333\).

Thus, \(P(1.98 < X < 2.03) = P(-1.333 < z < 0.333)\).

Using the standard normal distribution table, find \(\Phi(0.333)\) and \(\Phi(-1.333)\):

\(\Phi(0.333) = 0.6304\)

\(\Phi(-1.333) = 1 - \Phi(1.333) = 1 - 0.9087 = 0.0913\)

Therefore, \(P(-1.333 < z < 0.333) = \Phi(0.333) - (1 - \Phi(1.333)) = 0.6304 + 0.9087 - 1 = 0.539\).