Weird question. I checked with a computer, and all values are possible.

Yep, I can confirm all are possible. The smallest $n$ which give the digits for $0,1,2,...,9$ respectively are: $n=1,6,2,33,5,29,18,10,21,3$

Maybe the problem is for $\sqrt{n^4+2n^2+n}$?

@TuZo: The problem is stated as is in MOP 2005 booklet. Here is a slightly systematic way of reasoning. Note that, the first digit after the decimal point of $\sqrt{n^3+2n^2+n}$ is, $N \pmod{10}$ where $N=\lfloor 10(n+1)\sqrt{n}\rfloor$. Suppose that $N=10k+r$. This gives us, $$ 10k+r \leq 10\sqrt{n}(n+1)<10k+r+1 \implies k+\frac{r}{10}\leq \sqrt{n}(n+1)<k+\frac{r+1}{10}. $$Taking squares, we have $$ k^2+\frac{kr}{5}+\frac{r^2}{100}\leq n(n+1)^2 <k^2+\frac{k(r+1)}{5}+\frac{(r+1)^2}{100} = k^2+\frac{kr}{5}+\frac{r^2}{100}+\frac{kr}{5}+\frac{r}{50}+\frac{1}{100}. $$From here, it seems all $r$'s are achievable by suitable choice of $n,k$, via trial and error. However, in an exam setting I am not sure whether this is the best approach.