Answer
The radius of curvature of the lower surface of the lens is $1.00\,\mathrm{m}$.
Work Step by Step
Using the expression derived in the previous problem for the ring number:
$m={r^2\over \lambda R}-{1\over 2}$
For the nth ring, $m=n-1$, (since $m=0$ is the first ring)
$n-1={(0.162\times 10^{-2})^2\over 546\times 10^{-9} R}-{1\over 2}$
For the (n+20)th ring, $m=n+19$, and we have
$n+19={(0.368\times 10^{-2})^2\over 546\times 10^{-9} R}-{1\over 2}$
Subtracting the above two equations,
\begin{align*}
n+19-(n-1)&={(0.368\times 10^{-2})^2\over 546\times 10^{-9} R}-{1\over 2}-{(0.162\times 10^{-2})^2\over 546\times 10^{-9} R}+{1\over 2}\\
20&={(0.368\times 10^{-2})^2-(0.162\times 10^{-2})^2\over 546\times 10^{-9} R}\\
R&={0.109\times 10^{-4}\over 0.109\times 10^{-4}}\\
R&=1.00\,\mathrm{m}
\end{align*}
The radius of curvature of the lower surface of the lens is $1.00\,\mathrm{m}$.
