Let's take some values of n.
Clearly, the recurrence is $T(n)=T(logn)+1$
I'll take it as $T(n)=T\left \lceil (logn) \right \rceil+1$ because $logn = O\left \lceil (logn) \right \rceil$
n = $1024$
$1024\rightarrow 10\rightarrow 4\rightarrow 2$
3 recursive calls.
So, for 1024, it is $O(loglogn)$
n = $2^{2048}$
$2^{2048}\rightarrow 2048\rightarrow 12\rightarrow 4\rightarrow 2$
4 recursive calls.
So, for $2^{2048}$ it is $O(logloglogn)$
n = $2^{2^{2^{2048}}}$
$2^{2^{2^{2048}}}\rightarrow $ $2^{2^{2048}}\rightarrow $ $2^{2048}\rightarrow 2048\rightarrow 12\rightarrow 4\rightarrow 2$
6 recursive calls.
So, for $2^{2^{2^{2048}}}$ it is $O(loglogloglogn)$
It is evident that number of logs is dependent upon how big the input value is. So, it would be Option C