Answer
$R = 93~cm$
Work Step by Step
We can find the distance the ball falls in time $t = 0.65~s$:
$y = \frac{1}{2}~at^2$
$y = \frac{1}{2}(9.8~m/s^2)(0.65~s)^2$
$y = 2.07~m$
Thus the ball is at a height $s = 0.93~m$ above the mirror at this time.
Note that the image is also located at $s' = 0.93~m$
We can find the radius of curvature:
$\frac{1}{s} + \frac{1}{s'} = \frac{1}{f}$
$\frac{1}{s} + \frac{1}{s'} = \frac{2}{R}$
$\frac{1}{0.93~m} + \frac{1}{0.93~m} = \frac{2}{R}$
$\frac{2}{0.93~m} = \frac{2}{R}$
$R = 0.93~m$
$R = 93~cm$
