Here, object size, `h_(1) = +2.5 cm`, object distance, `u = -25 cm`

focal length of diverging mirror, `f = +20 cm`

image distance, `v = ?`, image size, `h_(2) = ?`

Using mirror formula, `1/u + 1/v = 1/f` or `1/v = 1/f - 1/u`

and putting, `u = -25 cm` and `f = +20 cm`, we get

`1/v = 1/20 -1/(-25) = (5+4)/100 = 9/100` or `v = 100/9 = 11.1 cm`

Since image is at the back of the mirror at 11.1 cm from the mirror, it must be virtual and erect. If `h_(2)` is the size of the image, then as

`m = h_(2)/h_(1) = (-v)/u`, `h_(2)/(2.5) = (-11.1)/(-25)` or `h_(2) = (11.1)/25 xx 2.5 = 1.11 cm`

This is the size of the image.