Given, the power of the lens, $P_1=-1.6 \mathrm{D}$ and $P_2=+2.1 \mathrm{D}$

$\because P=P_1+P_2=-1.6 \mathrm{D}+2.1 \mathrm{D}=0.5 \mathrm{D}$, now use the relation between focal length and power of the lens, 

$\therefore P=\frac{1}{f\left(\mathrm{m}\right)}\Rightarrow P=\frac{100}{f\left(\mathrm{cm}\right)}\Rightarrow 0.5=\frac{100}{f}$.

Focal length of the combination, $f=200 \mathrm{cm}$.