Let $G = U_{21}$.
Find the cyclic subgroup of $G$ generated by $[10]$.
I'm not sure how to do this but here is what I tried:
\begin{array}{l} g = 10 \equiv 10 \pmod {21} \\ g^2 = 100 \equiv 16 \pmod {21} \\ g^3 = 1000 \equiv 13 \pmod {21} \\ g^4 = 10000 \equiv 4 \pmod {21} \\ g^5 = 100000 \equiv 19 \pmod {21} \\ g^6 = 1000000 \equiv 1 \pmod {21} \end{array}
Once you get to $1$, the cycle repeats.
The subgroup has elements $[10], ???, [1] $.
$[10]$ has order $6$.
On
Are you sure the operation is multiplication? Since then, e.g. the element $3$ has no multiplicative inverse.
The operation is supposed to be addition modulo $21$ and in that case $10$ generates the whole $U_{21}$, since it is coprime to $21$ (being coprime is equivalent to the existence of $a$ and $b$ such that $a\cdot 21 + b \cdot 10 = 1$, so $b\cdot 10 = 1 \ mod\,21$ and since $1$ is evidently a generator, you can generate the whole group).
On
Indeed, you have found all the elements of $\langle 10\rangle \lt U_{21}$. $\langle 10 \rangle = \left\{[1], [4], [10], [13], [16], [19]\right\},\,$ and there are exactly six elements, so $\;\left|\langle 10 \rangle\right| = 6.\;$
A good "sanity" check when determining the order of a subgroup $H$ generated by an element $h \in G$ (for finite $G$) is using Lagrange's Theorem: If $H\leq G$, and $G$ is finite, then we know that the order of $H$ must divide the order of $G$.
In this case, $\left|U_{21}\right| = \varphi(21) = 12$, and sure enough, $6 \mid 12$. $\quad\large\checkmark$
To find out what $10^2$ is modulo $21$, take its remainder upon division by $21$. You'll find that this is $16$, so that $10^2=16$ in your group. Next, we have $10^3=10\cdot 16=160$. You'll find that the remainder of $160$ upon division by $21$ is $13$, so that $10^3=13$ in your group.
So far, we have the elements $1,10,16,13$. See if you can continue the reasoning above to find all of the elements in $\langle 10\rangle \subset U_{21}$.