Solution

In order for the image from the second lens to be a real image, the image from the first lens (which serves as the object for the second lens) has to be farther than $30$ cm (i.e. the focal length) in front of the second lens. This means that the (virtual) image from the first lens has to be farther than $15$ cm in front of the first lens. From the thin lens equation, the position of the object when the image is $15$ cm in front of the first lens is

$$u=\frac{v-f}{vf}=\frac{-15-30}{-15\cdot30}=-10.$$

Therefore, the object has to be closer than $10$ cm from the first lens.