The $\textsf{LL(1)}$ parsing table for the given grammar is:
$\begin{array}{|c|c|c|} \hline
& a&b&c \\ \hline
S& S\to aSa & S\to b & S \to c \\ \hline
\end{array}$
For any given input symbol $a,b$ or $c,$ the parser has a unique move from the start and the only state – so no conflicts.
As there is no conflict in $\text{LL(1)}$ parsing table, the given grammar is $\textsf{LL(1)}$ and since every $\textsf{LL(1)}$ is also $\textsf{LR(1)},$ the given grammar is $\textsf{LL(1)}$ as well as $\textsf{LR(1)}.$