Given that the probability that the time is greater than k is 0.675, we have:

\(P(X > k) = 0.675\)

Using the standard normal distribution, we find the corresponding z-value for the cumulative probability of \(1 - 0.675 = 0.325\).

The z-value for 0.325 is approximately \(z = -0.454\).

Using the standardization formula:

\(z = \frac{k - 140}{12}\)

Substitute \(z = -0.454\):

\(\frac{k - 140}{12} = -0.454\)

Solving for k:

\(k - 140 = -0.454 \times 12\)

\(k - 140 = -5.448\)

\(k = 140 - 5.448\)

\(k = 134.552\)

Rounding to two decimal places, \(k = 134.55\).