If $f(2n)$ is the number of ways of tiling a $2n\times 3$ rectangle with $2\times 1$ rectangles (with $f(0)=1$), then it's easy to derive the recurrence relation $f(2n+2)=f(2n)+2(f(2n)+f(2n-2)+\ldots+f(2)+f(0)),\ \forall n\ge 0$. This means that modulo $3$, the sequence $f(0),f(2),f(4),\ldots$ looks like this: $1,0,2,2,0,1,1,0,2,2,0,1,1,0,2,\ldots$ (it's $1,0,2$ followed by $2,0,1$, and then by $1,0,2$ again, and so on). In other words, $f(2n)\equiv 0\pmod 3$ whenever $n\equiv 1\pmod 3$ (so $f(200)\equiv 0\pmod 3$, since $200=2\cdot 100$ and $100\equiv 1\pmod 3$).