If ${\text{R}}_{\text{1}}{\text{ and R}}_{\text{2}}$  are the radii of curvature of double convex lens made of same material, the lens with more focal length is:

$\begin{array}{rcl}\frac{1}{\mathrm{f}}& =& (\mathrm{\mu }-1)(\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2})\\ \frac{1}{\mathrm{f}}& \propto & (\frac{1}{20}+\frac{1}{10})\propto \frac{3}{20}\\ \frac{1}{\mathrm{f}}& \propto & (\frac{1}{20}+\frac{1}{20})\propto \frac{1}{10}\\ \frac{1}{\mathrm{f}}& \propto & (\frac{1}{10}+\frac{1}{10})\propto \frac{2}{10}\\ \frac{1}{\mathrm{f}}& \propto & (\frac{1}{5}+\frac{1}{5})\propto \frac{4}{10}\end{array}$

Hence, the lens with more focal length will be the lens having radii as ${\text{R}}_{\text{1}}{\text{ = R}}_{\text{2}}\text{ = 20 cm.}$