You want to enforce $X(k) = 0 \implies R(k) = 0$ and $X(k) = 1 \implies R(k) \le G(k)$. You can use indicator constraints for that.
Alternatively, a straightforward big-M formulation yields
\begin{align}
R(k) &\le M_1(k) X(k) \tag1 \\
R(k) - G(k) &\le M_2(k) (1 - X(k)) \tag2 \\
\end{align}
A natural choice for $M_1(k)$ is a small constant upper bound $\bar{G}(k)$ on $G(k)$, and a natural choice for $M_2(k)$ is $\bar{G}(k) - 0$. But notice that you can do better because $M_2(k)$ needs to be an upper bound on $R(k)-G(k)$ only when $X(k)=0$; in that case, $(1)$ forces $R(k)=0$, so you can take $M_2(k)=0$:
\begin{align}
R(k) &\le \bar{G}(k) X(k) \tag1 \\
R(k) &\le G(k) \tag2 \\
\end{align}
Depending on other constraints in your model, you might be able to tighten further.