Double-convex lenses are to be manufactured from a glass of refractive index 1.55,

Refractive index of glass, $\mu=1.55$

Focal length of the double-convex lens, f = 20 cm

Radius of curvature of one face of the lens = R1

Radius of curvature of the other face of the lens = R2

Radius of curvature of the double-convex lens = R

$\therefore R_{1}=R$ and $R_{2}=-R$

The value of R can be calculated as:

$\frac{1}{f}=(\mu-1)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$

$\frac{1}{20}=(1.55-1)\left[\frac{1}{R}+\frac{1}{R}\right]$

$\frac{1}{20}=0.55 \times \frac{2}{R}$

$\therefore R=0.55 \times 2 \times 20=22 \mathrm{~cm}$

Hence, the radius of curvature of the double-convex lens is 22 cm.