Find the $1000$th digit after the decimal point of $\sqrt{n},$ where $n=\underbrace{11\dots1}_{1998 \text{ 1's}}$

A little experimentation shows that if $n$ is the integer with $2m$ digits, where $m$ is an integer, all of them $1,$ then $\sqrt{n}$ has $m$ $3$'s, followed by a decimal point, $m$ $3$'s, a $1,$ and $2m$ $6$'s. If this was true, that would imply that the required digit is $1.$ Just substitute $2m=1998$ to verify.

Let $x=\dfrac{10^m-\frac{1}{2}10^{-m}}{3},m\in\mathbb{N}.$ Then $x$ has $m$ $3$'s, followed by a decimal point, followed by $m$ $3$'s, followed by one $1,$ followed by infinite $6$'s. So we just need to show that $(x-10^{-m-2})^2<n$ and $(x+10^{-m-2})^2>n,$ where $n=\dfrac{10^{2m}-1}{9}.$ This will show that the $(m+1)$st digit, or $1000$th digit, is indeed $1.$

$$(x-10^{-m-2})^2=\left(\dfrac{10^m}{3}-\dfrac{47}{300}10^{-m}\right)^2\\ =\dfrac{10^{2m}}{9}-\dfrac{47}{450}+\dfrac{2209}{90\;000}10^{-2m}\\ =n-\dfrac{47}{450}+\dfrac{2209}{90\;000}10^{-2m}\\ <n$$.

Similarly, $$(x+10^{-m-2})^2=\left(\dfrac{10^m}{3}+\dfrac{47}{300}10^{-m}\right)^2\\ =\dfrac{10^{2m}}{9}+\dfrac{47}{450}+\dfrac{2209}{90\;000}10^{-2m}\\ =n+\dfrac{47}{450}+\dfrac{2209}{90\;000}10^{-2m}\\ >n$$.

It seems you actually are on the right track. Solving "smaller" problems of the "same kind" often pays off, and this is one of the times it does.

You already found that

$$ n = \frac{10^{1998}-1}{9} = \frac{10^{1998}}{9} - \frac19.$$

The binomial expansion for $(a+b)^{1/2}$ with $a = \frac{10^{2m}-1}{9}$ and $b = -\frac19$ gives us \begin{multline} \left(\frac{10^{2m}}{9} - \frac19\right)^{\!1/2} = \left(\frac{10^{2m}}{9}\right)^{\!1/2} + \frac12 \left(\frac{10^{2m}}{9}\right)^{\!-1/2} \left(-\frac19\right)\\ - \frac18 \left(\frac{10^{2m}}{9}\right)^{\!-3/2} \left(-\frac19\right)^{\!2} + \frac1{16} \left(\frac{10^{2m}}{9}\right)^{\!-5/2} \left(-\frac19\right)^{\!3} + \cdots.\end{multline}

Now try the following comparisons: \begin{align} \left(\frac{10^{2m}}{9}\right)^{\!1/2} && \text{vs.} &&& \dfrac{10^{m}-1}{3}+\dfrac13, \\ \end{align} \begin{align} \frac12 \left(\frac{10^{2m}}{9}\right)^{\!-1/2} \left(-\frac19\right) && \text{vs.} &&& - \dfrac16 10^{-m} \\ \end{align} \begin{align} - \frac18 \left(\frac{10^{2m}}{9}\right)^{\!-3/2} \left(-\frac19\right)^{\!2} + \frac1{16} \left(\frac{10^{2m}}{9}\right)^{\!-5/2} \left(-\frac19\right)^{\!3} + \cdots && \text{vs.} &&& 10^{-2m} \end{align}

You should be able to confirm the formula that you worked out from the pattern of digits in $\sqrt{11},$ $\sqrt{1111},$ and $\sqrt{111111}.$