Suppose $n$ is such that the decimal expansion of ${\large{\frac{1}{n}}}$ has exactly $2010$ digits after the decimal point.
Since the decimal expansion terminates, it follows that $n=2^a5^b$ where $a,b$ are nonnegative integers.
By hypothesis,
$$\frac{1}{n}=\frac{x}{10^{2010}}$$
where $x$ is a positive integer.
Note that $x$ can't be a multiple of $10$, else by removing a common factor of $10$ from the numerator and denominator of
$${\large{\frac{x}{10^{2010}}}}$$
it would follow that the decimal expansion of ${\large{\frac{1}{n}}}$ has less than $2010$ digits.
\begin{align*}
\text{Then}\;\;&\frac{1}{n}=\frac{x}{10^{2010}}\\[4pt]
\iff\;&n=\frac{10^{2010}}{x}\\[4pt]
\end{align*}
hence $x$ divides $10^{2010}$.
Thus, the only possible prime factors of $x$ are $2$ and $5$.
Since $x$ is not a multiple of $10$, it follows that $x$ can't have both $2$ and $5$ as prime factors, hence $x$ must be a power of $2$ or a power of $5$.
To minimize $n$, we want to maximize $x$.
The largest power of $2$ which divides $10^{2010}$ is $2^{2010}$, and the largest power of $5$ which divides $10^{2010}$ is $5^{2010}$.
It follows that the largest possible value of $x$ is $5^{2010}$, hence the smallest possible value of $n$ is $2^{2010}$.
Finally, note that $2^{2010}$ has exactly $2011$ positive integer divisors, namely$2^0,2^1,2^2,...,2^{2010}$.
