Answer
The lens should be located $~~2.0~m~~$ from the wall.
$f = 0.67~m$
Work Step by Step
Let $x$ be the distance of the lens from the wall.
We can find $x$:
$\vert M \vert = \frac{s'}{s} = 2$
$\frac{x}{3.0-x} = 2$
$x = 6.0-2x$
$3x = 6.0~m$
$x = 2.0~m$
The lens should be located $~~2.0~m~~$ from the wall.
Then $~~s' = 2.0~m~~$ and $~~s = 1.0~m~~$
We can find the focal length:
$\frac{1}{s}+\frac{1}{s'} = \frac{1}{f}$
$\frac{1}{1.0~m}+\frac{1}{2.0~m} = \frac{1}{f}$
$\frac{3}{2.0~m} = \frac{1}{f}$
$f = \frac{2.0~m}{3}$
$f = 0.67~m$
