Answer
7.2 cm past the second lens.
Work Step by Step
The first lens is converging, with f = 20.0 cm. An object placed at infinity forms an image at the focal point of the converging lens (see equation 23–8).
$$d_{i1}=f_1=20.0cm$$
This image is now the object for the second, diverging lens. This image is 6.0 cm beyond the second lens, so the object distance is negative.
$$d_{o2}=-6.0cm$$
Use the lens equation, 23–8, for the second lens.
$$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$
$$d_i=\frac{fd_o}{d_o-f}=\frac{(-36.5cm)(-6.0cm)}{-6.0cm-(-36.5cm)}=7.2cm$$
The final image is real, and is 7.2 cm past the second lens.