Let $f_{1} = 2^{2^{\sqrt{\log \log n}}}$
$f_{2} = 2^{\sqrt{\log n}}$
$f_{3} = n$
Assuming base $2$, $f_{2}$ can be rewritten as $2^{2^{log\sqrt{logn}}}$
$f_{3}$ can be rewritten as $2^{logn} = 2^{2^{loglogn}}$
As square root function has less growth rate than linear function,
$log\sqrt{logn} < loglogn \Rightarrow f_{2} < f_{3}$
$\sqrt{loglogn} < loglogn \Rightarrow f_{1} < f_{3}$
$f_{2}$ can also be rewritten as $2^{2^\frac{loglogn}{2}}$
$\sqrt{loglogn} < \frac{loglogn}{2}\Rightarrow f_{1} < f_{2}$
Hence $f_{1} < f_{2} < f_{3}$
Hence correct option is $A$.
Similar question: https://gateoverflow.in/333131/tifr2020-b-10