#18 Hua Luo Geng.

$\begin{aligned} n & = \left(19^3-{\color{#3D99F6}3\times 18 \times 19} - 1\right)^2 \\ & = \left(19^3-{\color{#3D99F6}3(19-1)(19)} - 1\right)^2 \\ & = \left(19^3-{\color{#3D99F6}3(19^2) + 3(19)} - 1\right)^2 & \small \color{#3D99F6} \text{Note that }(a-1)^3 = a^3-3a^2+3a-1 \\ & = \left(19 - 1\right)^{3 \times 2} \\ & = 18^6 = \left(2\times 3^2\right)^6 = 2^{\color{#D61F06}6} \times 3^{\color{#D61F06}12} \end{aligned}$

Therefore the number of factors of $n$ , $\sigma_0 (n) = ({\color{#D61F06}6} + 1)({\color{#D61F06}12}+1) = \boxed{91}$

Note that if $k$ is a positive integer, then $(k-1)^3=k^3-3k^2+3k-1=k^3-3k(k-1)-1=k^3-3\times k\times (k-1)-1$

From that $19^3-3\times 18\times 19-1=(19-1)^3=18^3$

$18=2^1\times 3^2$ . Ergo $(18^3)^2=(2^{1\times 3}\times 3^{2\times 3})^2=(2^3\times 3^6)^2=2^{3\times 2}\times 3^{6\times 2}=2^6\times 3^{12}$ .

Therefore $n$ has $(6+1)\times (12+1)=7\times 13=\boxed{91}$ divisors.