Given, 

Radius of curvature, R= 3 m

Object distance, u= -5 m

Let the image distance= v

and height of image = h¹

therefore,

Focal length, f =R/3 = 3/2 = 1.5 m

Using lens formula,

1/v +1/u = 1/f

1/v= 1/f-1/u 

= 1/1.5 -(-1/5) =0.867m

so, v= 1.15 m

The image is formed at a distance of 1.15 m behind the mirror.

Now, magnification,m = image distance/object distance = -v/u

= -1.15/(-5) = 0.23

So, the image is virtual and erect.