Left?

Entertaining problem... the presence of dominant terms makes it easier to handle.

Using a root of unity filter, we find $\sum_{k=1}^{672}{2015 \choose 3k-1}\approx \frac{2^{2015}}{3}=A$ . The error, due to neglecting the contributions of $\omega$ and $\omega^2$ , is miniscule: Less than 1 out of a number exceeding $10^{600}$ . Considering the derivative of $(1+x)^{2015}$ and using a root of unity filter once again, we find $\sum_{k=1}^{672}(3k-1){2015 \choose 3k-1}\approx \frac{2015*2^{2014}}{3}=B$ . Now $\sum_{k=1}^{672}k{2015 \choose 3k-1}\approx \frac{A+B}{3}=\frac{2017*2^{2014}}{9}$ . We use technology to see that the 21st digit from the left is a $\boxed{2}$