Answer
$r=\sqrt{(m+{1\over 2})\lambda R}$
Work Step by Step
For interference maximum to be observed the path difference between the rays reflecting off the curved surface of the lens and the flat surface of the glass plate must be:
$2d=(m+{1\over 2})\lambda$
Let $\theta$ be the angle of air wedge between the lens and the plate. Though $\theta$ changes along the wedge due to the curvature of the lens, we approximate it to be constant since the lens is assumed to have a very small curvature.
Then from geometry, $d=r\tan\theta\approx r\theta$
Since for small angles $\tan\theta\approx\theta$.
$\implies 2r\theta=(m+{1\over 2})\lambda$
But from geometry, as can be seen in the figure, $\tan2\theta={r\over\sqrt{R^2-r^2}}\implies 2\theta\approx {r\over\sqrt{R^2-r^2}}$. Substituting this in the above equation,
\begin{align*}
{r^2\over\sqrt{R^2-r^2}}&=\left(m+{1\over 2}\right) \lambda\\
{r^2\over R\sqrt{1-r^2/R^2}}&=\left(m+{1\over 2}\right) \lambda\\
{r^2\over R}&\approx \left(m+{1\over 2}\right) \lambda \,\,\,\,\,(\because r/R\ll 1)\\
r^2&=\left(m+{1\over 2}\right) \lambda R\\
r&=\sqrt{\left(m+{1\over 2}\right) \lambda R}
\end{align*}
