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

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.

If tis is what they were looking for in interview answer, then I must say it's a nasty question. There's no freaking way I'd remember this inequality. I'd remember that we studied it in Probability theory class, of course. However, these inequalities are so obtuse that nobody uses them in practice outside academic research, there's no place in my brain to store this junk. I'd never ask something like this on a job interview.

# "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
$$\sigma^2=M^2/2-(M/2)^2=M^2/4$$


  [1]: https://en.wikipedia.org/wiki/Popoviciu%27s_inequality_on_variances