Given observations are $x_1,x_2,...,x_{10}$

$\Rightarrow \sum _{i=1}^{10}\left(x_i-\alpha \right)=2$

$\Rightarrow \sum _{i=1}^{10}{x}_i-10\alpha =2$

Mean, $\mu =\frac{6}{5}=\frac{\sum x_i}{10}$

$\Rightarrow \sum x_i=12$

$\Rightarrow 10\alpha +2=12$

$\Rightarrow \alpha =1$

Now, $\sum _{i=1}^{10}{\left(x_i-\beta \right)}^2=40$. Let $y_i=x_i-\beta$

$\Rightarrow {\sigma }_y^2=\frac{1}{10}\sum y_i^2-{\left(\overline{y}\right)}^2$

$\Rightarrow \frac{84}{25}=\frac{1}{10}\sum {\left(x_i-\beta \right)}^2-{\left(\frac{{\displaystyle \sum _{i=1}^{10}\left(x_i-\beta \right)}}{10}\right)}^2$

$\Rightarrow \frac{84}{25}=4-{\left(\frac{12-10\beta }{10}\right)}^2$

$\Rightarrow {\left(\frac{6-5\beta }{5}\right)}^2=4-\frac{84}{25}$

$\Rightarrow {\left(\frac{6-5\beta }{5}\right)}^2=\frac{16}{25}$

$\Rightarrow 6-5\beta =\pm 4$

$\Rightarrow \beta =\frac{2}{5}$ (not possible) or $\beta =2$

Hence, $\frac{\beta }{\alpha }=2$