Area $A = l \times b = 16.2 \times 10.1 = 163.62\text{ cm}^2$. Relative error $\frac{\Delta A}{A} = \frac{\Delta l}{l} + \frac{\Delta b}{b} = \frac{0.1}{16.2} + \frac{0.1}{10.1} \approx 0.00617 + 0.00990 = 0.01607$. Absolute error $\Delta A = A \times 0.01607 = 163.62 \times 0.01607 \approx 2.63\text{ cm}^2$. Since the least significant figure in the measurements is at the tenths place and the product must follow the least significant figures of the factors (3 sig figs here), we round 163.62 to 164. The error is rounded to 3 to match the precision. Thus, $164 \pm 3$ is the most appropriate representation according to NCERT guidelines for propagation and rounding.