Explanation

If $\Delta x$ is error in a physical quantity, then relative error is calculated as $\Delta x/x$. Given that, Length $\ell = (16.2 \pm 0.1)cm$. Breadth $b = (10.1 \pm 0.1)cm \because \Delta l = 0.1cm$, $\Delta b = 0.1cm$ Area $(A) = l \times b = 16.2 \times 10.1 = 163.62cm^{2}$. In significant figure rounding off to three significant digits, area $A = 164cm^{2}$ $\Delta A/A = \Delta l/l + \Delta b/b = 0.1/16.2 + 0.1/10.1 = 2.63/163.62$ So, $\Delta A = A \times (2.63/163.62) = 164 \times (2.63/163.62) = 2.636cm^{2}$ Now rounding off up to one significant figure $\Delta A = 3cm^{2}$. So, Area $A = A \pm \Delta A = (164 \pm 3)cm^{2}$.