The unit of the rate constant is $\text{molar per second}$ ($\text{M s}^{-1}$), which indicates that it is a zero-order reaction. For a zero-order reaction, the integrated rate equation is: $[A]_t = [A]_0 - kt$ The amount of reactant $A$ consumed is $kt$. Since the stoichiometry of the reaction is $1:1$, the concentration of product $B$ formed will be equal to the amount of $A$ consumed. $[B] = kt$ Given data: $k = 0.6 \times 10^{-3} \text{ M s}^{-1}$ $t = 20 \text{ min} = 20 \times 60 \text{ s} = 1200 \text{ s}$ Substituting the values into the equation: $[B] = (0.6 \times 10^{-3} \text{ M s}^{-1}) \times 1200 \text{ s}$ $[B] = 0.6 \times 1.2 \text{ M} = 0.72 \text{ M}$ Therefore, the concentration of $B$ after $20 \text{ minutes}$ is $0.72 \text{ M}$.