If you have upper and lower bounds M and 0, then you can apply Popovivicu's upper bound on variance: $$\sigma^2<\frac 1 4M^2$$

Once you have the variance, apply usual sample mean distribution logic, i.e. the variance of a sample mean $\bar x$ to be $\sigma^2_{\bar x}\sim\sigma^2/n$. The rest is trivial.

$$\frac {10}{(\sigma^2_{\bar x}/n)}=2.58$$ $$n=0.258\sigma^2_{\bar x}=0.067M^2$$

"Derivation"

Ok, how would I proceed to derive this? I'd imagine the distribution with highest variance (entropy) possible, that is bounded between M and 0. That must be Bernoulli: $$x=0:p=1/2$$ $$x=M:p=1/2$$ Then the mean is $$\mu=M/2$$ and the variance is $$\sigma^2=M^2/2-(M/2)^2=M^2/4$$