The reason why the error is approximately the first neglected term has to do with the theory of power series and their convergence, an undergraduate topic that you aren't meant to know: hence the question telling you to assume it.

You already know that $\sqrt{1-x} \approx 1 - \frac{x}{2} - \frac{x^2}{8}$ with the next term being $-\frac{x^3}{16}$.

You use the first three terms to evaluate $\sqrt{11}$ by first evaluating $\sqrt{\frac{99}{100}} = \frac{3}{10}\sqrt{11}$.

So the error in evaluating $\frac{3}{10}\sqrt{11}$ is $-\frac{x^3}{16}$ with $x=\frac{1}{100}$.

But that means the error in evaluating $\sqrt{11}$ is the error in evaluating $\frac{3}{10}\sqrt{11}$ divided by $\frac{3}{10}$. After all, the multiply a quantity that has an error attached multiples the error along with it. Hence the necessity for division here.

This works out to be $$\displaystyle \frac{-\frac{x^3}{16}}{3/10} = -\frac{10}{3}\cdot \frac{x^3}{16} \, \mathrm{with} \, x=\frac{1}{100}$$

I'm not sure why you're using $8000/24000$ instead of just $10/3$ or why you think it's 'approximate'. You need to divide by exactly $10/3$ to recover the approximate error in $\sqrt{11}$ from knowing the approximate error in $\frac{3}{10}\sqrt{11}$. I hope this makes sense, if not - please let me know and I'll do my best to clarify.