Answer C : $\rightarrow F_1<F_3<F_2$
$\rightarrow$ comparing $ F_1 $and $F_2$
$F_1(n)=n^{\sqrt n}$ ,
$F_2(n)=2^n$
let, $n=K^2$
$F_1(K^2) = (K^2)^{\sqrt{K^2}} = (K^2)^K=K^{2K} =2^{2Klog_2(K)}$
$F_2(K^2)=2^{(K^2)}$
$K^2 >2Klog_2(K)$
$ \therefore F_2 > F_1$
$\rightarrow$ comparing $ F_1 $and $F_3$
$F_1(K^2)=2^{2Klog_2(K)}$
$F_3(K^2)=2^{(\frac {K^2}{2})}.(K^2)^{10}$
here simply, $\frac {K^2}{2}>2.Klog_2(K)$
$ \therefore F_3 > F_1$
$\rightarrow$ comparing $ F_2 $and $F_3$
$F_2=2^n=2^{\frac n 2}\times 2^{\frac n 2}$
$F_3=2^n=2^{\frac n 2}\times n^{10}$
$2^{\frac n 2}>n^{10}$ as $exponentiation > polynomial$
$ \therefore F_2 > F_3$
Thanks @
Anuj parashar for rectifying mistake.
Alternate way to compare $F_1$ and $F_2$
taking log on both sides,
$log_2(F_1(n))=\sqrt n log_2(n)$
$log_2(F_2(n))=n log_2(2)=n$
now putting $n=K^2$,
$log_2(F_2(n))=K^2$
$log_2(F_1(n))=\sqrt {K^2}log_2(K^2)=2Klog(K)$
$K^2>2Klog_2(K)$