3|\geq 28$. Since $28=(4-1)^3+1$, we can use Theorem \[kflemma\] to conclude that $\cI_3$ contains a $4$-flower. Let $k\geq 4$ be maximum such that $\cS$ is a $k$-flower in $\cI_3$, and let $C$ be the core of $\cS$. As $\cI_3$ is $3$-uniform and intersecting, every subfamily $\cG\sse \cI$ has $\tau(\cG)\leq 3$, which implies that $C\neq \mt$. Suppose first that $C=\{a\}$, and suppose $\cI$ is not a star centered at $a$. Let $A\in \cI$ be such that $a\notin A$. Consider the family $\cS_C$. As $\tau(\cS_C)\geq 4$, there exists some $S_1\in \cS_C$ such that $A\cap S_1=\mt$. Consequently, if $S^\pr=S_1\cup \{a\}$, then $S^\pr\in \cI$ and $A\cap S^\pr=\mt$, a contradiction. As a result, we may assume that $C=\{a,b\}$. This implies that $\cS_C$ is a family of singletons. Consequently, $\cS$ is a sunflower with at least $4$ petals.[^5] Additionally, for every $A\in \cI_3$, $A\cap \{a,b\}\neq\mt$.\
Let $\cA=\{A\in \cI_3:A\cap C=\{a\}\}$, and let $\cB=\{B\in \cI_3:B\cap C=\{b\}\}$. We have $|\cI_3|=|S|+|\cA|+|\cB|$. Let $\cAp=\{A-\{a\}:A\in \cA\}$, and $\cBp=\{B-\{b\}:B\in \cB\}$. If $\cAp=\mt$ or $\cBp=\mt$, we can conclude that $\cI_3$, and hence, $\cI$ is a star (centered at either $a$ or $b$), so suppose both are non-empty. Since $\cI$ is intersecting, $\cAp$ and $\cBp$ are cross-intersecting families, i.e. for any $A\in \cAp$ and $B\in \cBp$, $A\cap B\neq \mt.$ Let $V(\cAp)$ and $V(\cBp)$ be the vertex sets of $\cAp$ and $\cBp$ respectively, and let $n(\cX)=|V(\cX)|$ for $\cX\in \{\cAp,\cBp\}$. We first prove the following claims.

\[clm1\] If both $\cAp$ and $\cBp$ are intersecting, or $|\cAp|\geq 2$ and $|\cBp|\geq 2$, then, $|\cX|\leq 2+ n(\cX)$ for each $\cX\in \{\cAp,\cBp\}$.

If $\cAp$ is intersecting, it is either a triangle, or a star. In either case, the bound follows trivially. A similar argument works for $\cBp$, so suppose, without loss of generality that $\cAp$ has two disjoint edges, say $\{xy,x^\pr y^\pr\}$. $\cBp\sse \{xy^\pr,y^\pr y,yx^\pr,x^\pr x\}$, giving the required bound for $\cBp$. Now, if $\c