jmc

We can write $$\begin{array}{rl}1061520150601& =1\cdot 100^6+6\cdot 100^5+15\cdot 100^4\\ & +20\cdot 100^3+15\cdot 100^2+6\cdot 100+1.\end{array}$$Notice that the cofficients on powers of 100 are all binomial. In fact, we have $$\begin{array}{rl}1061520150601& =\left(\genfrac{}{}{0ex}{}{6}{6}\right)\cdot 100^6+\left(\genfrac{}{}{0ex}{}{6}{5}\right)\cdot 100^5+\left(\genfrac{}{}{0ex}{}{6}{4}\right)\cdot 100^4\\ & +\left(\genfrac{}{}{0ex}{}{6}{3}\right)\cdot 100^3+\left(\genfrac{}{}{0ex}{}{6}{2}\right)\cdot 100^2+\left(\genfrac{}{}{0ex}{}{6}{1}\right)\cdot 100+\left(\genfrac{}{}{0ex}{}{6}{0}\right).\end{array}$$By the binomial theorem, this is equal to $(100+1{)}^6$, so its sixth root is $\boxed{101}$.