Worst case running time of lexicographical sorting of a list of n strings each of length n using merge sort
The mistake and correction

$$T(n)= 2T(n/2) + \theta(n^2)$$

As you intended, $T(n)$ is the worst time of mergesorting $n$ strings of length $n$. Then, $T(n/2)$ on the RHS means the worst time of mergesorting $n/2$ strings of length $n/2$. So during the mergesort, the length of strings shrinks from $n$ to $n/2$!

What should have been done is using a different variable to express the number of strings at the start of each mergesort. Let us define $T(m, n)$ to be the cost of mergesorting $m$ strings of length $n$. Now, we will have

$$T(m, n)= 2T(m/2, n) + \theta(mn),$$

from which we can deduce $$T(m,n) = \theta(nm\log m).$$



Another correct approach

Mergesort of $n$ items takes $\theta(n\log n)$ time, assuming a unit cost of comparison. In fact, mergesort uses $\theta(n\log n)$ comparisons. Now that it costs $\theta(n)$ to compare two strings of length $n$ at worst time, the worst time of mergesort of $n$ strings of length $n$ should be $\theta(n\cdot n\log n)$.