For a chemical reaction A $$\rightarrow$$ B, it was found that concentration of B is increased by 0.2 mol L$$^{-1}$$ in 30 min. The average rate of the reaction is _________ $$\times 10^{-1}$$ mol L$$^{-1}$$ h$$^{-1}$$. (in nearest integer)

For the reaction $$\mathrm{A \rightarrow B}$$ we are given that the concentration of product $$\mathrm{B}$$ increases by $$0.2\;\text{mol L}^{-1}$$ during a time interval of $$30\;\text{min}$$.

First, recall the definition of average rate of a reaction:

$$\text{Average rate} = \frac{\text{Change in concentration}}{\text{Time interval}}$$

Here, the change in concentration of $$\mathrm{B}$$ is

$$\Delta [\mathrm{B}] = +0.2\;\text{mol L}^{-1}.$$

The given time interval is $$30\;\text{min}$$. Because the final unit required is $$\text{mol L}^{-1}\,\text{h}^{-1}$$, we must convert minutes to hours. We have

$$30\;\text{min} = \frac{30}{60}\;\text{h} = 0.5\;\text{h}.$$

Now we substitute these values into the rate formula:

$$\text{Average rate} = \frac{0.2\;\text{mol L}^{-1}}{0.5\;\text{h}} = 0.4\;\text{mol L}^{-1}\,\text{h}^{-1}.$$

Next, we express $$0.4$$ in scientific notation with a power of $$10^{-1}$$ so that it matches the required format "_____ × 10-1 $$\text{mol L}^{-1}\,\text{h}^{-1}$$". We write

$$0.4 = 4 \times 10^{-1}.$$

Thus, the blank should be filled by the integer $$4$$.

Hence, the correct answer is Option 4.