Though $\sqrt{n}=2^m$ is the shortest method.
$T(n)=2T(n^{1/2}) +1$
$=2 *[ 2 T((n^{1/2})^{1/2}) +1] +1$
$=2^{2} T((n^{1/4}) +2 +1$
$=2^{3} T((n^{1/2^{3}}) +4 +2 +1$
$=2^{k} T((n^{1/2^{k}}) +2^{k-1}+ 2^{k-2}+..............+4 +2 +1$
$=2^{k} T((n^{1/2^{k}}) +[(2^{k}-1)/ 2-1] * 2^{0}$
as per given termination condition is $n^{1/2^{k}} =2$
$\Rightarrow log(n^{1/2^{k}}) =log(2)$
$\Rightarrow log(n) =2^{k}$
$\Rightarrow log( log(n)) =k$
inserting value of k or 2^k in above equation
$T(n)= log(n) * T(2) + log(n) - 1$
$T(n)= 3 log(n) - 1$
$T(n)= \Theta (log(n))$
Answer is B