We should put the object closer to the lens. To calculate this let us start with the lens equation

$$
frac{1}{f}=frac{1}{d_0}+frac{1}{d_i}.
$$
This rearranged gives

$$
frac{1}{d_i} = frac{1}{f}-frac{1}{d_0} = frac{d_0-f}{d_0f}
$$
which finally becomes

$$
d_i=frac{d_0f}{d_0-f}.
$$
The magnification is

$$
m=-frac{d_i}{d_0}
$$
which can be rewritten as

$$
m=-d_itimesfrac{1}{d_0} text{ (substitute now the derived formula for $d_i$) } m=-frac{f}{d_0-f}.
$$
Now we have to solve this for $d_0$. This is done in the following sequence

$$
f-d_0=frac{f}{m}Rightarrow d_0=f-frac{f}{m}=f-frac{f}{-3} = frac{4}{3}f= 6.7text{ cm}.
$$

If the object is placed between the lens and the focal point on a convex lens, in this case the image is $textbf{virtual}$ and $textbf{upright}$.

If the object is placed between the lens and the focal point on a convex lens, in this case the image is $textbf{virtual}$ and $textbf{upright}$.

If the object is placed between the lens and the focal point on a concave lens, in this case the image is $textbf{virtual}$ and $textbf{upright}$.

If the object is placed between the lens and the focal point on a concave lens, in this case the image is $textbf{virtual}$ and $textbf{upright}$.

Using the lens equation we have

$$
frac{1}{f}=frac{1}{l_0}+frac{1}{l_i}.
$$
This gives

$$
frac{1}{l_i}=frac{1}{f}-frac{1}{l_0}Rightarrow frac{1}{l_i}=frac{l_0-f}{fl_0}.
$$
This finally yields for the image distance

$$
l_i=frac{fl_0}{l0-f}=frac{-32text{ cm}times 23text{ cm}}{23text{ cm}-(-32text{ cm})}=-13.38text{ cm}.
$$

Now by definition the magnification is

$$
m=-frac{l_i}{l_0}=-0.58.
$$