Let \(X\) be the lifetime of the light bulb. We know \(X \sim N(2000, \sigma^2)\).

We are given \(P(X > 1800) = 0.96\). This can be rewritten in terms of the standard normal variable \(Z\) as:

\(P\left(Z > \frac{1800 - 2000}{\sigma}\right) = 0.96\).

Using the standard normal distribution table, we find that \(P(Z > -1.751) = 0.96\), so:

\(\frac{200}{\sigma} = 1.751\).

Solving for \(\sigma\), we get:

\(\sigma = \frac{200}{1.751} \approx 114\).