What's $n$ ?

The sum is:

$n(1+2+\cdot+99)=n\times \dfrac{99\times 100}2=\dfrac{11n}{2}\times (900)$

Now we know $900$ is a perfect square then for whole number to be a perfect square $\dfrac{11n}{2}$ must be a perfect square and it happens if we arrange one $11$ in numerator for completing its pair and arrange a $2$ in numerator for taking care of denominator we are done. Hence $n=11\times 2$ . $n^2=484$ , which implies number of digits in $n^2$ is $3$ .

Consider

$\begin{aligned} S & = n + 2n + 3n + \cdots + 99n \\ & = \frac {(99)(100)n}2 \\ &= \color{#D61F06}{2}\cdot 3^2 \cdot 5^2 \cdot \color{#D61F06}{11} \cdot n \end{aligned}$

The smallest perfect square $S$ is one with the least number of squared prime factors . The remaining two prime factors not squared are 2 and 11, therefore $n = 2\cdot 11$ to make $S=\color{#D61F06}{2^2}\cdot 3^2\cdot 5^2\cdot \color{#D61F06}{11^2}$ . Therefore $n^2 = 2^2\cdot 11^2 = 484$ which has $\boxed{3}$ digits.