jmc

If $x-3$ is a factor of $f(x)=x^3-3x^2+tx+27$, then using the Factor Theorem, we know that $f(3)=0$. We have $$\begin{array}{rl}f(3)& =3^3-3(3^2)+t(3)+27\\ & =27-27+3t+27\\ & =3t+27.\end{array}$$So $3t+27=0$. We can solve this to get $t=\boxed{-9}$.