#### Answer

$9$ inches

#### Work Step by Step

The work required to compress the spring beyond equilibrium can be calculated as:
$\text{Work}, W= \int_{m}^{n} kx \ dx $; where $k$ is the spring constant.
and $\dfrac{1}{2} k(1.5)^2=18 \implies k =16\ lb/ft$
Now, $F=k x \implies 12 =16 x $
or, $x=0.75 \ ft$
Thus, a weight of $ 12 \ lb$ will stretch the spring by $9$ inches.