= Solution
Here $H$ is the <paw graph>, a triangle with a pendant edge.
For the lower bound, partition the vertices of $K_6$ into two triples. Colour the edges inside each triple red and all edges between the triples blue. Each red component is only a triangle, while the blue graph is the triangle-free graph $K_{3,3}$, so there is no monochromatic copy of $H$. Hence $R(H)>6$.
For the upper bound, every colouring of $K_7$ contains a monochromatic triangle because $R(3)=6$. Suppose that a triangle $T$ is red, and let $S$ be the other four vertices. If any edge from $T$ to $S$ were red, it would be a pendant edge extending $T$ to a red $H$. Thus all edges between $T$ and $S$ are blue.
If an edge $xy$ inside $S$ were blue, then for any $u\in T$ the vertices $u,x,y$ would form a blue triangle, and an edge from a second vertex of $T$ to $x$ would extend it to a blue $H$. Therefore every edge inside $S$ is red. The resulting red $K_4$ contains a red triangle and an additional incident edge, hence a red $H$. The blue-triangle case is symmetric, so the <Ramsey number of the paw graph> is
$$
\boxed{R(H)=7}.
$$
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