Show that $R$ a subring of $M_3(\mathbb{R})$ 
Let $R =\left \{  \begin{pmatrix}
0 &a  &b \\ 
 0& 0 & c\\ 
 0&  0& 0
\end{pmatrix}:a,b,c\in \mathbb{R}\right \}$ . Show that $R$ a subring of $M_3(\mathbb{R})$. Is it a ring with identity? If so, is it a unital subring.

Clearly, the zero matrix lies in $R$. Also,
$$\begin{pmatrix}
0 &a_1  &b_1 \\ 
 0& 0 & c_1\\ 
 0&  0& 0
\end{pmatrix}- \begin{pmatrix}
0 &a_2  &b_2 \\ 
 0& 0 & c_2\\ 
 0&  0& 0
\end{pmatrix}= \begin{pmatrix}
0 &a_1-a_2  &b_1-b_2 \\ 
 0& 0 & c_1-c_2\\ 
 0&  0& 0
\end{pmatrix} \in R $$
and
$$\begin{pmatrix}
0 &a_1  &b_1 \\ 
 0& 0 & c_1\\ 
 0&  0& 0
\end{pmatrix}  \begin{pmatrix}
0 &a_2  &b_2 \\ 
 0& 0 & c_2\\ 
 0&  0& 0
\end{pmatrix}= \begin{pmatrix}
0 &0  &a_1c_2 \\ 
 0& 0 & 0\\ 
 0&  0& 0
\end{pmatrix}\in R $$
Is it a ring with identity? I do not think so, since the product of these matrices are a subset of $R$ and thus, there is no matrix $I$, such that $AI=IA=A$ for all $A\in R$. Am I right?
Finally, whether or not there is an identity matrix, we have that $R$ is not a unital subring. Since identity in $M_3(\mathbb{R})$ is  $\begin{pmatrix}
1&0  &0 \\ 
 0& 1 & 0\\ 
 0&  0& 1
\end{pmatrix}\notin R$.

Your reasons for the first and third parts are fine, but not the second:

Is it a ring with identity? I do not think so, since the product of these matrices are a subset of  and thus, there is no matrix , such that == for all ∈. Am I right?

Merely being a subset of $R$ does not preclude a subring from having its own identity. The identity $e$ of any ring always satisfies $e^2=e$, that is, it is idempotent.  In fact, if you have any such element $e\in R$, then $eRe$ is always a subring with identity $e$.
But let's keep going with this fact that $e^2=e$. Note that in the ring in question, $x^3=0$ for every element $x$.  It should be obvious to you now that nothing is going to satisfy $e^2=e$ except $0$, which is clearly not the identity.
