Step 1: Work with volumes:
Balanced reaction: $2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2$.
Using the molar volume at STP, 2 mol $\text{KClO}_3$ give $3\times 22.4 = 67.2$ L of oxygen.
Step 2: Unitary method:
For $67.2$ L of $\text{O}_2$ we need 2 mol of $\text{KClO}_3$.
For $11.2$ L we need $2\times\frac{11.2}{67.2}$ mol.
\[ 2\times\frac{11.2}{67.2} = \frac{2}{6} = \frac{1}{3}\text{ mol} \]
So the answer is option (B).
Final Answer:
The required amount is $\frac{1}{3}$ mol.
\[ \boxed{\frac{1}{3}\text{ mol}} \]