Answer
The force required is $20$ pounds.
Work Step by Step
Consider that the distance by which the spring will stretch is D and the force required to stretch the spring is F. It is provided that the variable D varies directly as F, that is $D\ =\ kF$.
Thus,
$\begin{align}
& 9=12k \\
& k=\frac{9}{12} \\
& k=\frac{3}{4}
\end{align}$
Substitute D = 15 and $k=\frac{3}{4}$ in $D\ =\ kF$.
$\begin{align}
& D=kF \\
& 15=\frac{3}{4}F \\
& F=\frac{15\times 4}{3} \\
& F=20
\end{align}$
Thus, a force of 20 pounds is required to stretch the spring by 15 inches.
