For an associative ring \(A\), define \(A^{[1]}\) to be \(A\) and \(A^{[n]}\) (\(n>1\)) to be the two-sided ideal of \(A\) that is generated by all \(n\)- fold Lie commutators \([a_ 1,[a_ 2,\dots,[a_{n-1},a_ n]\dots]]\) (\(a_ i\in A\)). \(A\) is called Lie-nilpotent if \(A^{[n]}=0\) for some \(n\), in which case the smallest such \(n\) is denoted \(t_ L(A)\). Defining ideals \(A^{(n)}\) of \(A\) inductively by \(A^{(1)}=A\), \(A^{(n)}=[A^{(n-1)},A]A\) and putting \(t^ L(A)=\min\{n: A^{(n)}=0\}\), one has \(A^{(n)}\subseteq A^{[n]}\) for all \(n\) and thus \(t^ L(A)\geq t_ L(A)\). In the special case where \(A=KG\) is the group algebra of a group \(G\) over a field \(K\), \textit{I. B. S. Passi}, \textit{D. S. Passman} and \textit{S. K. Sehgal} [Can. J. Math. 25, 748-757 (1973; Zbl 0266.16011)] have shown that \(KG\) is Lie-nilpotent precisely if either \(\text{char} K=0\) and \(G\) is abelian or \(\text{char} K=p>0\) and \(G\) is nilpotent with \([G,G]\) a finite \(p\)-group. Thus, in the former case, one has \(t_ L(KG)=t^ L(KG)=2\). In general, \(t^ L(KG)\) can be computed from the Lie-dimension subgroups of \(G\). The main result of the present article shows that if \(KG\) is Lie-nilpotent and \(\text{char} K=p>3\) then \(t^ L(KG)=t_ L(KG)\).