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Since the maximum value of $y=a{x}^2+bx+c$ is 5, which occurs at $x=3$, this tells us that the vertex of the parabola is $(3,5)$. Hence, the quadratic is of the form $y=a(x-3{)}^2+5$, where $a$ is a negative number. (We know that $a$ is negative because $y$ has a maximum value.)

We are also told that the graph passes through the point $(0,-13)$. Substituting these coordinates into the equation $y=a(x-3{)}^2+5$, we get $-13=9a+5$, so $a=(-5-13)/9=-18/9=-2$. Therefore, the equation is $y=-2(x-3{)}^2+5$.

When $x=4$, we have $m=-2\cdot 1^2+5=\boxed{3}$.