A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 93-cm and a standard deviation of 0.8-cm.

To find the probability that the length of a randomly selected steel rod is between 92.3 cm and 93 cm, we can use the properties of the normal distribution. We first need to convert the lengths into z-scores.

The formula for a z-score is given by:

\[
z = \frac{(X - \mu)}{\sigma}
\]

where \( X \) is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

Given:
- Mean \( \mu = 93 \, \text{cm} \)
- Standard deviation \( \sigma = 0.8 \, \text{cm} \)

Now, we calculate the z-scores for 92.3 cm and 93 cm.

1. For \( X = 92.3 \, \text{cm} \):

\[
z_{92.3} = \frac{(92.3 - 93)}{0.8} = \frac{-0.7}{0.8} = -0.875
\]

2. For \( X = 93 \, \text{cm} \):

\[
z_{93} = \frac{(93 - 93)}{0.8} = \frac{0}{0.8} = 0
\]

Next, we will find the probabilities corresponding to these z-scores using the standard normal distribution table or a calculator with normal distribution functions.

- The probability for \( z = 0 \):

\[
P(Z < 0) = 0.5
\]

- The probability for \( z = -0.875 \):

Using a z-table or a normal distribution calculator:

\[
P(Z < -0.875) \approx 0.1894
\]

Now, we can find the probability that the length of a randomly selected steel rod is between 92.3 cm and 93 cm:

\[
P(92.3 < X < 93) = P(Z < 0) - P(Z < -0.875) = 0.5 - 0.1894 = 0.3106
\]

Thus, the final probability is:

\[
\boxed{0.3106}
\]