= Solution
A <finite field> $\mathbb F_q$ contains a primitive tenth root exactly when $10\mid(q-1)$. Its multiplicative group is cyclic: the exponent of a finite abelian <subgroup> of a field is attained by an element, by combining its prime-power components; since every element is a root of $X^e-1$, the <polynomial> root bound forces the group order to be at most that exponent, hence equal to it. A cyclic group contains an element of order ten exactly when ten divides its order.
Over $\mathbb F_q$, the conjugates of $\xi$ are its <Frobenius automorphism> orbit $\xi,\xi^q,\xi^{q^2},\ldots$. The extension degree is the least $d>0$ with $q^d\equiv1\pmod{10}$. For $q=3$ this order is four, for $q=11$ it is one, and for $q=19$ it is two. Therefore
$$
\boxed{\mathbb F_3(\xi_{10})\cong\mathbb F_{81},\qquad
\mathbb F_{11}(\xi_{10})=\mathbb F_{11},\qquad
\mathbb F_{19}(\xi_{10})\cong\mathbb F_{361}.}
$$
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