Let $(a,b,c)\in\mathbb Z_{+}^3$ then basically, you have the following system of equation :
$$\begin{cases}n=20a\\
n^2=b^3 \\
n^3=c^2\end{cases}$$
Since
$$n=b\sqrt b=\frac {c}{\sqrt [3]c}\tag 1$$
then we have $b=b_1^2$ and $c=c_1^3$, where $(b_1,c_1)\in\mathbb Z_{+}^2$ . This implies,
$$n=b_1^3=c_1^2\tag 2$$
Thus, we find that $n=2^2\cdot 5\cdot a$ must be both a perfect square and a perfect cube . Therefore,
$$a=5p^2=2\cdot 5^2\cdot q^3\tag 3$$
where $(p,q)\in\mathbb Z_{+}^2$ .
This leads to the following :
$$\begin{align}&p^2=2\cdot 5\cdot q^3\\
\implies &p=q\sqrt {2\cdot 5\cdot q}\\
\implies &q=2\cdot 5\cdot q_1^2,\thinspace q_1\in\mathbb Z_{≥1}\end{align}\tag 4$$
Finally, we conclude that :
$$\begin{align}\min \{q\}&=10\\
\min\{p\}&=100\end{align}\tag 5$$
which implies,
$$\begin{align}&a=5p^2=5\cdot 10^4\\
\implies &n=20a=10^6\thinspace.\end{align}\tag 6$$
Thus, the number of digits of $n$ is $6+1=\color{#c00}{7}$ .