jmc

Using the Carmichael function, we have $\lambda (1000)=100$, so $3^{100}=1(\mathrm{mod}1000)$. Therefore, letting $N=3^{3^3}$, we seek to find an $n$ such that $N\equiv n(\mathrm{mod}100)$ so that $3^N\equiv 3^n(\mathrm{mod}1000)$. Using the Carmichael function again, we have $\lambda (100)=20$, so $N=3^{27}\equiv 3^7(\mathrm{mod}100)\equiv 87(\mathrm{mod}100)$. Therefore $n=87$, and so we have the following:$$3^{3^{3^3}}\equiv 3^{87}(\mathrm{mod}1000).$$ Now, $$\begin{array}{rl}{3}^{87}=(3^{20}{)}^4\cdot 3^7& \equiv 401^4\cdot 187(\mathrm{mod}1000)\\ & \equiv 601\cdot 187(\mathrm{mod}1000)\\ & \equiv \boxed{387}(\mathrm{mod}1000).\end{array}$$