Hint: Use the mirror formula.

Step 1: Find the image of the ends of the object.

Since the object distance is u. Let us consider the two ends of the object to be at distance ${\mathrm{u}}_1=\mathrm{u}-\mathrm{L}/2 \mathrm{and} {\mathrm{u}}_2=\mathrm{u}+\mathrm{L}/2$, respectively so that $\left|{\mathrm{u}}_1-{\mathrm{u}}_2\right|=\mathrm{L}$. Let the image of the two ends be formed at ${\mathrm{v}}_1 \mathrm{and} {\mathrm{v}}_2$, respectively so that the image length would be
                        $\mathrm{L}\text{'}=\left|{\mathrm{v}}_1-{\mathrm{v}}_2\right|$                  ...(i)
Applying mirror formula, we have,
                                                   $\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}} \mathrm{or} \mathrm{v}=\frac{\mathrm{fu}}{\mathrm{u}-\mathrm{f}}$
On solving, the positions of two images are given by,
                                                          ${\mathrm{v}}_1=\frac{\mathrm{f} (\mathrm{u}-\mathrm{L}/2)}{\mathrm{u}-\mathrm{f}-\mathrm{L}/2}, {\mathrm{v}}_2=\frac{\mathrm{f}(\mathrm{u}+\mathrm{L}/2)}{\mathrm{u}-\mathrm{f}+\mathrm{L}/2}$

Step 2: Find the length of the image.
For length, substituting the value in (i), we have,
                                                           $\mathrm{L}\text{'}=\left|{\mathrm{v}}_1-{\mathrm{v}}_2\right|=\frac{{\mathrm{f}}^2\mathrm{L}}{{(\mathrm{u}-\mathrm{f})}^2-{\mathrm{L}}^2/4}$
Since, the object is short and kept away from the focus, we have.
                                                             ${\mathrm{L}}^2/4<<{(\mathrm{u}-\mathrm{f})}^2$
Hence, finally,                                 $\mathrm{L}\text{'}=\frac{{\mathrm{f}}^2}{{(\mathrm{u}-\mathrm{f})}^2}\mathrm{L}$
This is the required expression of the length of the image.