The mean and standard deviation of $100$ observations were calculated as $40$ and $5.1,$ respectively by a student who took by mistake $50$ instead of $40$ for one observation, then the correct standard deviation is

Standard deviation $\left(\sigma \right)=\sqrt{\frac{1}{n}\sum _{i=1}^n{x}_i^2-\frac{1}{n^2}{\left(\sum _{i=1}^n{x}_i\right)}^2}$
$=\sqrt{\frac{1}{n}\sum _{i=1}^n{x}_i^2-{\left(\stackrel{-}{x}\right)}^2}$

 $\Rightarrow 5.1=\sqrt{\frac{1}{100}\times \mathrm{I}\mathrm{n}\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\sum _{i=1}^n{x}_i^2-{\left(40\right)}^2}$
or, $26.01=\frac{1}{100}\times$ Incorrect $\sum _{i=1}^n{x}_i^2-1600$
Therefore,
Incorrect $\sum _{i=1}^n{x}_i^2=100(26.01+1600)=162601$

Correct mean, $\overline{x }=\frac{3990}{100}=39.9$
Now, Correct $\sum _{i=1}^n{x}_i^2=$ Incorrect $\sum _{i=1}^n{x}_i^2-{\left(50\right)}^2+{\left(40\right)}^2$
$=162601-2500+1600=161701$
Therefore correct standard deviation
$=\sqrt{\frac{\mathrm{C}\mathrm{o}\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\sum x_i^2}{n}-{\left(\mathrm{C}\mathrm{o}\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{}\mathrm{m}\mathrm{e}\mathrm{a}\mathrm{n}\right)}^2}$
$=\sqrt{\frac{161701}{100}-{\left(39.9\right)}^2}=\sqrt{1617.01-1592.01}=\sqrt{25}=5$