$G$ is a graph with exactly two blocks $B_1$ and $B_2$. $B_1$ and $B_2$ share a single vertex. Can you express the chromatic number of $G$ as a function of the chromatic number of $B_1$ and $B_2$?
Note: it is not specified whether G is connected or not
Let $X(G)$ represent the chromatic number of $G$
I think the answer is: $X(G) \leq X(G - B1 - B2 ) + 2$
I got this because I am pretty sure that a graph that is just two blocks connected by a single vertex would have a chromatic number of $2$, so we remove $B_1$ and $B_2$ from $G$ and add back $2$.
here is another post from this website that will probably help: Prove that the chromatic number of a graph is the same as the maximum of the chromatic numbers its blocks.
We know that there cannot exist an edge from $B_1$ to $B_2$ as this form a cycle and $G$ would be 2-connected itself. By this, saying $G-B_1-B_2$ doesn't make much sense as it will be an empty graph.
We now note that $X(G)= \max \{X(B_1), X(B_2) \}$ as we can colour $B_1$ and $B_2$ independent of each other.