Ans: It is given that,
Distance of object in front of mirror, $u=-10cm$ (negative sign due to the location of the object in front of the mirror)
Distance of image from mirror, $v=?$
It is known that,
Magnification of a spherical mirror,$m=\frac{{{h}_{i}}}{{{h}_{o}}}=-\frac{v}{u}$
where,
${{h}_{i}}$ is the height of the image
${{h}_{o}}$ is the height of the object
Let, ${{h}_{o}}=h$
It is given that three times the enlarged real image of the object is produced.
So, ${{h}_{i}}=-3h$ ($-$ due to real image formation)
$\Rightarrow m=-\frac{3h}{h}=-\frac{v}{u}$
$\Rightarrow 3=\frac{v}{-10}$
$\Rightarrow v=-30cm$(negative sign due to the formation of inverted image)
Therefore, the location of image from the mirror is at a distance of $30cm$ and the nature of the image is inverted.