GIven confidence level,11.24<u<12.01 and that n=90 + c= .90, find the sample standard devaiton o
I'm assuming that since you ask for sample standard deviation rather than population standard deviation, the confidence interval has been constructed using the t-distribution:
$\displaystyle L_1 = \bar{x} - \frac{s}{\sqrt{n}} \, t_{n-1, \, \alpha/2} < \mu < \bar{x} + \frac{s}{\sqrt{n}} \, t_{n-1, \, \alpha/2} = L_2$
Therefore $\displaystyle L_2 - L_1 = \frac{2 s}{\sqrt{n}} \, t_{n-1, ~ \alpha/2}$.
In your case $\displaystyle \alpha = 0.1 \Rightarrow \frac{\alpha}{2} = 0.05$ (I think ..... it's not really clear from what you posted), n = 90, $\displaystyle L_1 = 11.24$ and $\displaystyle L_2 = 12.01$. Use a table of critical values to get $\displaystyle t_{89, ~ 0.05}$.
Substitute all these values and solve for s.