Applying one step of Richardson's Extrapolation

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Suppose $N(h)$ is an approximation for every $h>0$ and that $M=N(h)+Ah^{4/3}+Bh^{7/3}$. Apply one step of Richardson's Extrapolation with Value $N(3h)=2.7172$ and $N(h)=2.7182$ and compute the impoved value for $M$. If needed take $h=.0001$.

I'm not really sure how to proceed, any help or insight would be appreciated.

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In general, when you are approximating a quantity $M$ by a formula like $$ M = N(h) + A h^{k_0} + B h^{k_1}, \quad k_1>k_0 $$

the extrapolation based on $N(h)$ and $N(h/t)$ is given by $$ R(h,t)=\dfrac{t^{k_0} N(h/t)-N(h)}{t^{k_0}-1} $$

This corresponds, as it was pointed out in one of the comments to use the "M" equations and eliminate the constant $A$.

In your case, the extrapolated value would be $$ \frac{3^{4/3}N(h) - N(3h)}{3^{4/3}-1}\approx 2.7185 $$