$T(n) = \sqrt{n}\,T(\sqrt{n}) + n\log n$
At level 0, the top level, you'll have $1$ array, of size $n$.

At level 1, you'll have $n^{1/2}$ arrays for each array at level $0$, for a total of $n^{1/2}$ arrays, each of size $n^{1/2}$.

At level 2 you'll have $n^{1/4}$ arrays for each array at level 1 for a total of $n^{1/2}n^{1/4}$ arrays, each of size $n^{1/4}$.

At level 3 you'll have $n^{1/8}$ arrays for each array at level 2 for a total of $n^{1/2}n^{1/4}n^{1/8}$ arrays, each of size $n^{1/8}$.

In tabular form we have for the first few levels,
$$\begin{array}{ccccc}
\mathbf{level} &amp; \mathbf{arrays} &amp; \mathbf{size} &amp; \mathbf{contribution/array} &amp; \mathbf{total\ contribution}\\
0 &amp; 1 &amp; n &amp; n\lg n &amp; (1)(n\lg n)=n\lg n\\
1 &amp; n^{1/2} &amp; n^{1/2} &amp; n^{1/2}\lg n^{1/2} &amp; (n^{1/2})(n^{1/2}\lg n^{1/2})=n\lg n^{1/2}\\
2 &amp; n^{1/2}n^{1/4} &amp; n^{1/4} &amp; n^{1/4}\lg n^{1/4} &amp; (n^{1/2}n^{1/4})(n^{1/4}\lg n^{1/4})=n\lg n^{1/4}\\
3 &amp; n^{1/2}n^{1/4}n^{1/8} &amp; n^{1/8} &amp; n^{1/8}\lg n^{1/8} &amp; (n^{1/2}n^{1/4}n^{1/8})(n^{1/8}\lg n^{1/8})=n\lg n^{1/8}
\end{array}$$
So the total runtime, $T(n)$, will be the sum of the contributions at each level, namely
$$
n\lg n + n\lg n^{1/2} + n\lg n^{1/4} + n\lg n^{1/8} +\dotsb
$$
Use a property of the log function and you'll see that this simplifies to a simple expression.