If $(\lg n)^{2^{100}} < {n^{1/2}}$, where $\lg$ is the binary logarithm, then
$$(\lg n)^{2^{101}} < n$$ $$2^{101}\lg \lg n < \lg n$$ $$101 < \lg \lg n - \lg \lg \lg n$$
I don't know that whether I assume that $n = 2^x$. Anyway suppose that $n = 2^x$, then $\lg \lg \lg n = \lg \lg x$.
So, suppose that $x = 2^y$ (it's ambiguous), then $101 < y - \lg y$ therefore, $y = 108$, $n = 2^{2^{108}}$.
I'm curious that there is exact solving method. Because I'm not sure, if $n = 2^{2^{108}} - 1$, what happens??
One such $n$ will be:
$$n = 2^{2^{2^{7}}}$$
If you substitute the $n$ in the inequality you have shown:
$$(\lg{n})^{2^{101}} < n$$
by this number, then after some reductions you'll get a correct inequality below:
$$108 < 128$$
The number $2^{2^{2^{7}}}$ doesn't look like a minimal number with such property. Finding a minimal such number is another and much more difficult problem.