= Solution
Put $\alpha=44^{1/11}$ and $L=\mathbb Q(\alpha)$. The <Eisenstein criterion> at $11$ shows that $X^{11}-44$ is an <irreducible polynomial>, so $[L:\mathbb Q]=11$. Its derivative is $11X^{10}$, and the <polynomial discriminant> is
$$
\operatorname{disc}(X^{11}-44)=-11^{11}44^{10}=-2^{20}11^{21}.
$$
The <discriminant-index formula for an integral lattice> implies that the <field discriminant> has no prime divisor outside $\{2,11\}$. By the <discriminant obstruction to ramification>, every other prime is unramified.
To prove that neither candidate disappears through an index correction, let $\mathfrak p$ lie above $p\in\{2,11\}$ and normalize its <valuation> to have value group $\mathbb Z$. If $e=e_{\mathfrak p/p}$, the relation $\alpha^{11}=44$ gives
$$
11v_{\mathfrak p}(\alpha)=e\,v_p(44)=\begin{cases}2e,&p=2,\\e,&p=11.\end{cases}
$$
In both cases $11\mid e$. The <fundamental identity for prime decomposition>, $\sum_{\mathfrak p\mid p}e_{\mathfrak p/p}f_{\mathfrak p/p}=11$, now forces exactly one <prime ideal> above each candidate, with $e=11$ and $f=1$. This is <ramification forced by a coprime root valuation>; at $2$ it avoids an invalid direct application of the <Eisenstein criterion>, since $4\mid44$. Hence
$$
\boxed{\text{The ramified rational primes are exactly }2\text{ and }11;\ \text{both are totally ramified}.}
$$
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