Solution

Firstly, the magnification of the object is given by:
$k = \frac{s'}{s}$ (1)
where $s'$ is the image-mirror distance and $s$ is the object-mirror distance, with $k$ = 3.
And the relation for these distances is
$\frac{1}{s} +\frac{1}{s'} = \frac{1}{f}$ (2)
When object is moved a distance $d$ away from the mirror, the magnification changes
$m = \frac{s''}{s+d}$ (3)
with $m$ = 0.5.
And the relation for these distances is
$\frac{1}{s+d}+\frac{1}{s''}=\frac{1}{f}$ (4)
From (1) and (2), eliminating term $s'$,
$\frac{s}{f} = 1+\frac{1}{k}$ (5)
From (3) and (4), eliminating term $s''$,
$\frac{s}{f}=1+\frac{1}{m}-\frac{d}{f}$ (6)
Equaling (5) and (6), and separating $f$,
$f = \frac{mkd}{k-m} = 48\;\rm{cm}$