If $\Delta x$ is error in a physical quantity, then relative error is calculated as $\frac{\Delta x}{x}$.
Given that,
Length $l=(16.2 \pm 0.1) \mathrm{cm}$
Breadth $b=(10.1 \pm 0.1) \mathrm{cm} \quad \because \Delta l=0.1 \mathrm{~cm}, \Delta \mathrm{b}=0.1 \mathrm{~cm}$
Area $(A)=l \times b=16.2 \times 10.1=163.62 \mathrm{~cm}^2$
In significant figure rounding off to three significant digits, area $A=164 \mathrm{~cm}^2$
$$
\begin{aligned}
\frac{\Delta A}{A} &=\frac{\Delta l}{l}+\frac{\Delta b}{b}=\frac{0.1}{16.2}+\frac{0.1}{10.1} \\
&=\frac{1.01+1.62}{16.2 \times 10.1}=\frac{2.63}{163.62}
\end{aligned}
$$
So, $\begin{aligned} \Delta A=A \times & \frac{2.63}{163.62}=164 \times \frac{2.63}{163.62} \\=2.636 \mathrm{~cm}^2 \end{aligned}$
Now rounding off up to one significant figure $\triangle A=3 \mathrm{~cm}^2$.
So, Area $A=A \pm \triangle A=(164 \pm 3) \mathrm{cm}^2$.