$$908.80 = 250(1+\frac{0.24}{m})^{6m}$$
I get stuck here because one of the m's is part of the ln:
$$ln908.8 = m(6ln250 + 6ln250\frac{0.24}{m})$$
Not sure how to pull it out. I've tried various things but keep getting the wrong answer; I know the final answer is supposed to be m=0.99. Any help?
You will need the Lambert-$W$ function for this. There is no solution in terms of elementary functions.
EDIT: Rewrite the equation as
$${\left(\sqrt[6]{\frac{908.80}{250}}\right)}^{1/m}=1+0.24\frac1m$$
Letting $1/m=x$, we have
$${\left(\sqrt[6]{\frac{908.80}{250}}\right)}^{x}=1+0.24x$$
which is of the form $p^x=ax+b$ and whose solution is in general given by
$$x=\frac{-W\left(-\frac{\ln p}{a}\,p^{-b/a}\right)}{\ln p}-\frac{b}a$$
and making the approipriate substitutions yields $m$ in terms of the Lambert-$W$ function.
Without the W function you are in for a numeric solution. I like fixed point iteration, so would take logs $$908.80 = 250(1+\frac{0.24}{m})^{6m}\\\log 908.80=\log 250+6m\log(1+\frac {0.24}m)\\m=\frac{\log 908.80-\log 250}{6 \log(1+\frac{0.24}m)}$$ and iterate $$m_{i+1}=\frac{\log 908.80-\log 250}{6 \log(1+\frac{0.24}{m_i})}$$ to convergence at about $m=0.999968$
As Ross Millikan answered, a numerical method is required. So, considering that me need to find the zero of equation $$f(m)=\log (250)+6m\log\left(1+\frac {0.24}m\right)-\log( 908.80)$$ that is to say $$f(m)=m\log\left(1+\frac {0.24}m\right)-\frac16 \log\left(\frac {908.8}{250}\right)$$ let me be very lazy and start Newton iterations using $m_0=0.5$. The successive iterates will be $$\left( \begin{array}{cc} n & m_n \\ 0 & 0.50000000 \\ 1 & 0.78190010 \\ 2 & 0.95860749 \\ 3 & 0.99848163 \\ 4 & 0.99996602 \\ 5 & 0.99996794 \end{array} \right)$$
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