Answer:
[tex]1.0336 m[/tex]
Explanation:
A representative diagram with sign convention has been shown below. Assuming that mirror is thin having the focal length [tex]f[/tex] and the radius of curvature [tex]R[/tex], so that [tex]R=2f \; \cdots (i)[/tex].
Here, the object side is negative and the image side is positive as give that image is behind the mirror.
Let [tex]u[/tex] and [tex]v[/tex] be the object and image distance respectively, so [tex]u=-20.3 m[/tex]and [tex]v=50.4 cm =0.504 m[/tex].
Now, from the mirror formula
[tex]\frac 1 v + \frac 1 u =\frac 1 f[/tex]
Putting the given values in the above formula, we have
[tex]\frac 1 {50.4} + \frac 1 {0.504} =\frac 1 f \\\Rightarrow \frac 1 f=1.935 \\\Rightarrow f=0.5168 m[/tex]
Now, from the equation [tex](i)[/tex]
[tex]R=2 \times 0.5168 m \\\Rightarrow R= 1.0336 m.[/tex]
Hence, the radius of curvature of the mirror is [tex]1.0336 m[/tex].
