A convex lens of refractive index 1.55, with both the surfaces of the same radius of curvature has a focal length of 20 cm. The radius of curvature of the surface will be:

22 cm

$\frac{1}{f} = (\mu -1)(\frac{1}{R_1} -\frac{1}{R_2}) = (\mu -1)(\frac{1}{R} -\frac{1}{-R}) = (\mu -1)\frac{2}{R}$

$ R = 2(\mu -1)f = 2\times 0.55 \times 20 cm = 22cm$