The radius of a thin wire is 0.16 mm. The area of cross-section of the wire in mm² with correct number of significant figures is 

$$\begin{gathered} R=0.16 mm \\ \mathrm{Hence}, A= \mathrm{\pi }{R}^2 =\frac{22}{7}\times {\left(0.16\right)}^2=0.080384 \end{gathered}$$

Since, radius has two significant figures so answer also will have two significant figures.

$\therefore A=0.080$