$T(n) = \sqrt{n}*T(\sqrt{n}) + n$
Dividing both sides by $n$, we get
$\frac{T(n)}{n} = \frac{T(\sqrt{n})}{\sqrt{n}} + 1$
Now, let $S(n) = \frac{T(n)}{n}$
So, the recurrence equation boils down to
$S(n) = S(\sqrt{n}) + 1$
Let, $n = 2^{k}$
So, $S(2^{k}) = S(2^{\frac{k}{2}}) + 1$
Again, let $H(k) = S(2^{k})$
So, $H(k) = H(k/2) + 1$
Solving the equation by master theorem, we get
$S(2^{k}) = H(k) = logk$
Now, since $n = 2^{k}$, $k = logn$
So, $S(n) = loglogn$
Since, $S(n) = \frac{T(n)}{n}$,
$T(n) = nS(n)$
$T(n) = nloglogn$