Suppose that the average of $s$ and $t$ is $90$ or more; then $\frac{s+t}2\ge 90$, so $s+t\ge 180$, and therefore $r+s+t\ge r+180$. On the other hand, you know that $r+s+t=270$, so $270\ge r+180$, and therefore $90\ge r$. Is it possible to have $90\le r and $r+s+t=270$?
Added: You ask whether this argument is correct:
since $s+t = 270 -r$ so $\frac{270-r}{2}> 90$ thus $270-r > 180$ so $90-r >0$ Now since $r$ cant be negative which is impossible (even if it is) the $90-r > 0$
It needs a bit of work. The fact that $s+t=270-r$ does not imply that $\frac{270-r}2>90$, so your first so is inappropriate. What you want to say is this:
Suppose that $\frac{270-r}2>90$; then $270-r>180$, so $90-r>0$.
Next, it’s true that $r$ can’t be negative, but this is irrelevant. What you can conclude here is that $r<90$.
Unfortunately, none of this really helps: it just shows that there is no immediate problem with the assumption that the average of $s$ and $t$ is more than $90$. In order to show that this actually is the case, you want to suppose that it isn’t, and derive a contradiction. Thus, you really ought to start out:
Suppose that $\frac{270-r}2\le 90$; then $270-r\le 180$, so $90-r\le 0$, and therefore $90\le r$.
Now you can go on to argue as follows:
By hypothesis, $r, so we have $90\le r. This implies that $r+s+t>r+r+r\ge 3\cdot 90=270$ and hence that $\frac{r+s+t}3>\frac{270}3=90\;.$ This, however, contradicts the hypothesis that the arithmetic mean of $r,s$, and $t$ is $90$. Thus, the supposition that $\frac{270-r}2\le 90$ must be false, i.e., it must be true that $\frac{270-r}2>90$ and therefore that the arithmetic mean of $s$ and $t$ is more than $90$.