The given limit can be written as $$\begin{align}&\lim_{n\to\infty}\frac{96}{n^4}\sum_{m=1}^nm\sum_{k=1}^{n-m+1}k \\\\&=\lim_{n\to\infty}\frac{96}{n^4}\sum_{m=1}^{n}m\cdot\frac{(n-m+1)(n-m+2)}{2} \\\\&=\lim_{n\to\infty}\frac{48}{n}\sum^{n}_{m=1}\frac mn\bigg(1-\frac mn+\frac 1n\bigg)\bigg(1-\frac mn+\frac 2n\bigg) \\\\&=\lim_{n\to\infty}\frac{48}{n}\sum_{m=1}^{n}\frac mn\bigg(\bigg(1-\frac mn\bigg)^2+\frac 3n\bigg(1-\frac mn\bigg)+\frac{2}{n^2}\bigg) \\\\&=\lim_{n\to\infty}\frac{48}{n}\sum_{m=1}^{n}\frac mn\bigg(1-\frac mn\bigg)^2 \\&\qquad +\lim_{n\to\infty}\frac{48}{n}\sum_{m=1}^{n}\frac mn\bigg(\frac 3n\bigg(1-\frac mn\bigg)+\frac{2}{n^2}\bigg) \\\\&=\lim_{n\to\infty}\frac{48}{n}\sum_{m=1}^{n}\frac mn\bigg(1-\frac mn\bigg)^2 \\&\qquad +\lim_{n\to\infty}\frac{48}{n^4}\sum_{m=1}^{n}(3mn-3m^2+2m)\tag1\end{align}$$
The second term is equal to $0$, so we get $$\begin{align}(1)&=48\int_0^1x(1-x)^2dx \\\\&=48\int_0^1(x-2x^2+x^3)dx \\\\&=48\bigg(\frac 12-\frac 23+\frac 14\bigg) \\\\&=\color{red}4\end{align}$$
Added :
Ivan Kaznacheyeu's comment is helpful.
First of all, we have $$\lim_{n\to\infty}\frac{96}{n^4}\sum_{m=1}^{n}m\sum_{k=1}^{n-m+1}k=\lim_{n\to\infty}\frac{96}{n^4}\sum_{m=1}^{n}m\sum_{k=1}^{\color{red}{n-m}}k\tag1$$ because $$\begin{align}&\lim_{n\to\infty}\frac{96}{n^4}\sum_{m=1}^{n}m\sum_{k=1}^{n-m+1}k \\\\&=\lim_{n\to\infty}\frac{96}{n^4}\sum_{m=1}^{n}m\bigg(n-m+1+\sum_{k=1}^{\color{red}{n-m}}k\bigg) \\\\&=\underbrace{\lim_{n\to\infty}\frac{96}{n^4}\sum_{m=1}^{n}m(n-m+1)}_{=0} \\&\qquad +\lim_{n\to\infty}\frac{96}{n^4}\sum_{m=1}^{n}m\sum_{k=1}^{\color{red}{n-m}}k\end{align}$$
Now, $(1)$ can be written as $$(1)=96\lim_{n\to\infty}\frac{1}{n}\sum_{m=1}^{n}\frac mn\underbrace{\bigg(\frac 1n\sum_{k=1}^{n-m}\frac kn\bigg)}_{S}\tag2$$
Here, $S$ is as follows :
Therefore, we finally get $$\begin{align}(2)&=96\int_{0}^{1}x\bigg(\int_{0}^{1-x}y\ dy\bigg)dx \\\\&=96\int_0^1\frac{x(1-x)^2}{2}dx \\\\&=4\end{align}$$