Question 921625
You need to make {{{M=3}}} with a lens that has {{{f=6}}} .
Your formula is wrong. It should be {{{M=f/(f-p)}}} .
That way with numerator (f) and denominator (f-p) in the same units, you get a ratio of lengths for M.
(I was once a high school physics teacher, and I also checked the formula to make sure my memory was not failing me after so many years of not teaching this stuff).
Regardless of the formula, all you had to do was plug in the given values for M and f, and then solving for p.
Plugging those values in you get
{{{3<=6/(6-p)}}
Solving, ou get
{{{3<=6/(6-p)}} 
{{{3(6-p)<=6}}
{{{3*6-3p=6)}}
{{{18-3p<=6)}}
{{{18-6-3p<=0)}}
{{{12-3p<=0)}}
{{{12<=3p)}}
{{{12/3<=p)}}
{{{highlight(p>=4)}}}
The object be placed at most {{{highlight(4cm)}}} from the lens for its image to appear at least three times as large.