Given $\alpha , \beta$ are the roots of the equation $x^2+x+1=0.$

The given equation is of the form ${\omega }^2+\omega +1=0,$ which is the condition for $\omega \& {\omega }^2$ to be the imaginary cube roots of unity.

We know that, if $\omega$ is the cube root if unity, then ${\omega }^3=1,$ hence, ${\alpha }^3={\beta }^3=1$ also the sum and product of the roots of the quadratic equation are respectively $\alpha +\beta =-1, \alpha \beta =1.$

Now, $(2-\alpha )(2-\beta )\left(2-{\alpha }^{10}\right)\left(2-{\alpha }^{20}\right)=\left[(2-\alpha )(2-\beta )\right]\left[\left(2-{\left({\alpha }^3\right)}^3\alpha \right)\left(2-{\left({\alpha }^3\right)}^6{\alpha }^2\right)\right]$

$=\left[4-2\alpha -2\beta +\alpha \beta )\right]\left[\left(2-\alpha \right)\left(2-{\alpha }^2\right)\right]$

$=\left[4-2\left(\alpha +\beta \right)+\alpha \beta )\right]\left[4-2\left(\alpha +{\alpha }^2\right)+{\alpha }^3\right]$

$=\left[4-2\left(-1\right)+1\right]\left[4-2\left(-1\right)+1\right]$

$=7\times 7=49$