(a) Given that the volume of the volumetric flask is $V = 100 \text{ cm}^3 = 0.1 \text{ dm}^3$, the molarity of the solution is $S = 0.1 \text{ M}$, and the molar mass of $\text{Na}_2\text{CO}_3$ is $M = 106 \text{ g/mol}$. The required mass $W$ is calculated as: $$ \begin{aligned} W &= \frac{S \times M \times V}{1000} \text{ (if V is in cm}^3) \\ W &= \frac{0.1 \times 106 \times 100}{1000} \text{ g} \\ &= 1.06 \text{ g} \end{aligned} $$ Thus, $1.06 \text{ g}$ of $\text{Na}_2\text{CO}_3$ is required. (b) Given the mass of $\text{Na}_2\text{CO}_3$ taken is $W = 1.28 \text{ g}$, the volume of the solution is $V = 100 \text{ cm}^3$, and the molar mass is $M = 106 \text{ g/mol}$. The molarity $C$ is calculated as: $$ \begin{aligned} C &= \frac{W \times 1000}{M \times V} \\ &= \frac{1.28 \times 1000}{106 \times 100} \text{ M} \\ &\approx 0.12075 \text{ M} \end{aligned} $$ Rounding to two decimal places, the concentration is $0.12 \text{ M}$.