Notice that $a = 201320132013$ and $10^{12}$ are coprime. By Euler's theorem
$$a^{\phi\left(10^{12}\right)} \equiv 1 \quad \bmod \left(10^{12}\right)$$
Let $b = a^{\phi\left(10^{12}\right) - 1}$. We have $a b \equiv 1 \bmod \left(10^{12}\right)$. Because $b$ and $10^{12}$ are coprime, again by Euler's theorem
$$a \equiv a b^{\phi\left(10^{12}\right)} \equiv b^{\phi\left(10^{12}\right) - 1} \quad \bmod \left(10^{12}\right) .$$
But
$$\phi\left(10^{12}\right) - 1 = 2^{13} \times 5^{11} - 1 \equiv (-1)^{13} \times (-1)^{11} - 1 \equiv 0 \bmod (3) .$$
Therefore, there exists a positive integer $m$ such that $\phi\left(10^{12}\right) - 1 = 3 m$.
Let $n = b^{m}$. We have
$$n^{3} = b^{\phi\left(10^{12}\right) - 1} \equiv 201320132013 \bmod \left(10^{12}\right) .$$