It looks like $\Bbb{R}(n)$ is your notation for the $n\times n$ matrix algebra, so I'll use $R$ to denote that ring. Here's an outline solution that contains useful exercises about simple Artinian rings.
Step 1: Show that any simple right $R$-module $S$ is isomorphic to a simple right ideal of $R$. This can be done by noticing there is a map $\phi:R\rightarrow S$. Since $R$ is semisimple, $R=\ker(\phi)\oplus N$, and so $S\cong R/\ker(\phi)\cong N\subset R$.
Step 2: Show that if $S$ and $T$ are simple right ideals of a ring, then $ST=0$ or $S\cong T$. Since the ring $R$ is simple, $ST=0$ can't happen, because that would imply that $0\neq ann(S)\lhd R$. Conclude that $R$ has only one type of simple right module up to isomorphism.
Step 3: Show that the module you specified in your post is simple over $R$. You can do this, for example, by demonstrating that for any $x$ and any $y$ in $M$, there is a matrix $A$ such that $xA=y$.