jmc

We begin by reducing the factors of the product modulo 20: $$\begin{array}{rl}77& \equiv -3(\mathrm{mod}20),\\ 88& \equiv 8(\mathrm{mod}20),\\ 99& \equiv -1(\mathrm{mod}20).\end{array}$$(Note that we could have used the more "standard" reductions $77\equiv 17$ and $99\equiv 19$, but the reductions above will make our computations easier.)

Now we have $$\begin{array}{rl}77\cdot 88\cdot 99& \equiv (-3)\cdot 8\cdot (-1)\\ & =24\\ & \equiv \boxed{4}(\mathrm{mod}20).\end{array}$$