Since the image is 3 times the object magnification can be $\pm 3$

and focal length $=-6 \;cm$

We know $m= \large\frac{f}{f-u}$

For real image $m=-3$

$-3= \large\frac{-6}{-6-u}$

$u= -8 \;cm$

For virtual image $m=+3$

$3= \large\frac{-6}{-6-u}$

$u=-4 \;cm$

Hence a is the correct answer.