A particular fruit's weights are normally distributed, with a mean of 380 grams and a standard deviation of 8 grams.

To determine the weight above which 3% of the sample means will fall when picking 20 fruits, we can use the properties of the normal distribution.

1. **Identify the parameters of the distribution**:
- Population mean (\(\mu\)) = 380 grams
- Population standard deviation (\(\sigma\)) = 8 grams
- Sample size (\(n\)) = 20 fruits

2. **Calculate the standard deviation of the sample mean (\(\sigma_{\bar{x}}\))**:
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{8}{\sqrt{20}} \approx 1.7889 \text{ grams}
\]

3. **Find the z-score corresponding to the top 3% tail of the distribution**:
Since we want the mean weight greater than a certain value, we need to find the z-score that corresponds to 97% of the distribution (as the top 3% is in the right tail).

You can look up the z-value for 0.97 in the standard normal (z) table or use a calculator:
\[
z \approx 1.88079
\]

4. **Convert the z-score back to the corresponding weight**:
Using the formula for converting a z-score to an x-value:
\[
x = \mu + z \cdot \sigma_{\bar{x}}
\]
Plugging in the values:
\[
x = 380 + 1.88079 \cdot 1.7889 \approx 380 + 3.359 \approx 383.359
\]

5. **Round to the nearest gram**:
The value we calculated is approximately 383.359 grams, which rounds to 383 grams.

Therefore, 3% of the time, the mean weight of the 20 fruits will be greater than **383 grams**.