What is the object distance? you will need to use the magnification equation to find a relationship between s and s′. then substitute into the thin lens equation to solve for s.?

For lens in S I system, 1/v - 1 / u = 1/f

gives 1/v=1/f + 1/u,

multiply by 'u' on both sides

u/v = u/f + 1

u/v = (u+f) / f

inverting,

v/u = f / (u+f)

Therefore magnification 'm' = v/u=f / (u+f)

'm' = f / (u+f)

Given magnification is 5 / 9

f / (u+f) = 5 / 9

object distance 'u' = 4f / 5

Now substitute for focal length f (focal length of diverging lens is negative)