Radically Rousing

$\begin{aligned} \sqrt{\frac{4^{20}-2^{21}+1}{2^{20}+2^{11}+1}} & = \sqrt{\frac{\left(2^{20}\right)^2-2\left(2^{20}\right)+1}{\left(2^{10}\right)^2+2\left(2^{10}\right)+1}} = \sqrt{\frac{\left(2^{20}-1\right)^2}{\left(2^{10}+1\right)^2}} = \frac{2^{20}-1}{2^{10}+1} = \frac{(2^{10}-1)(2^{10}+1)}{2^{10}+1} = 2^{10}-1 \end{aligned}$

$\implies a = \boxed{10}$

Relevant wiki: Simplifying Expressions with Radicals - Intermediate

First we convert everything to powers of $2$ :

$\sqrt{\dfrac{4^{20}-2^{21}+1}{2^{20}+2^{11}+1}}=\sqrt{\dfrac{2^{40}-2^{21}+1}{2^{20}+2^{11}+1}}$

The powers of $2$ all seem to be multiples of $10$ or near to multiples of $10$ so let $x=2^{10}$ :

$\sqrt{\dfrac{2^{40}-2^{21}+1}{2^{20}+2^{11}+1}}=\sqrt{\dfrac{x^4-2x^2+1}{x^2+2x+1}}$

This suggests that $x^2+2x+1$ is a factor of $x^4-2x^2+1$ and checking we see it is:

$\sqrt{\dfrac{x^4-2x^2+1}{x^2+2x+1}}=\sqrt{\dfrac{(x^2+2x+1)(x^2-2x+1)}{(x^2+2x+1)}}=\sqrt{x^2-2x+1}$

$\sqrt{x^2-2x+1}=\sqrt{(x-1)^2}=x-1$

Substituting back in for $x$ we see the expression is equivalent to $2^{10}-1$ so $\color{#20A900}{\boxed{\boxed{a=10}}}$

of course the exact solution is straightforward, but there is an even more straightforward approximation rendering the same result: numerator is approximately 2^40, denominator approximately 2^20, so left side approximately 2^10, which gives a=10, approximately

(2^20-1)/(2^10+1)=2^10-1

We have to use two formulas which is $a^4-a^2b^2+b^4=(a^2-ab+b^2)(a^2+ab+b^2)$ and $(a-1)^2=a^2-2a+1$

$\sqrt{\frac{4^{20} - 2^{21} + 1}{2^{20}+2^{11}+1}} = \sqrt{\frac{2^{40} - 2^{21} + 1}{2^{20}+2^{11}+1}} = \sqrt{\frac{2^{21}(2^{19} - 1 + 2^{-21})}{2^{11}(2^{9}+1+2^{-11})}} =\sqrt{2^{10}\frac{2^{19} - 1 + 2^{-21}}{2^{9}+1+2^{-11}}}=2^5\sqrt{\frac{2^{19} - 1 + 2^{-21}}{2^{9}+1+2^{-11}}} \approx 2^5 \sqrt{\frac{2^{19}}{2^9}} = 2^5\sqrt{2^{10}} = 2^5\cdot2^5 = 2^{10} \approx 2^{10}-1$ .

If $a$ is an integer then $a=10$ .

After I typed this I read Lev's solution. Bravo for taking the path of least resistance! Although the factorization might be more straightforward, in general most things don't factor nicely, so I prefer to avoid that approach.