A \textit{Steiner triple} system of order \(n\) (or a triple system) is a pair \((S,T)\), where \(T\) is a collection of edge disjoint triangles, otherwise called triples, which partition the edge set \(K_n\) with vertex set \(S\). It is well known that the spectrum for Steiner triple systems is the set of all \(n \equiv 1\) or \(3 \pmod 6\) and that if \((S,T)\) is a triple system, \(| T| = n(n-1)/6\). The \textit{Hexagonal triple system} is the graph consisting of three triangles (triples) \(\{a,b,c\}\), \(\{c,d,e\}\) and \(\{e,f,a\}\), where \(a\), \(b\), \(c\), \(d\), \(e\), and \(f\) are all distinct. A \(\lambda\)-fold hexagonal triple system of order \(n\) is a pair \((X,H)\), where \(\lambda\) is a collection of edge disjoint hexagon triples which partitions the edge set of \(\lambda K_n\) with vertex set \(X\). The spectrum of three fold hexagonal triple systems is the set of all \(n\equiv 1\) or \(3\pmod 6\), and if \((X,H)\) is a three-fold hexagon triple systems then \(| H| = n(n-1)/6\). It can be noticed that the spectra of triple systems and three fold hexagonal triple systems are precisely the same. Also the number of hexagon triples in the three fold hexagon triple systems of order \(n\) is the same as the number of triples in Steiner triple systems. Then the question arises that for which \(n(n-1)/6\), it is possible to construct a three-fold hexagon triple system with the property that the inside triples form a Steiner triple system . \textit{Selda Küçükçifçi} and \textit{C.C. Lindner} [Discrete Math. 279, No.\,1-3, 325--335 (2004; Zbl 1043.05021); Des. Codes Cryptography 32, No.\,1-3, 251--265 (2004; Zbl 1053.05015)] proved that this as is always possible for all \(n \equiv 1\) or \(3 \pmod 6\). Later Lucia proved that every Steiner triple system is the inside of some three-fold hexagon triple system. Considering the spectrum of one-fold hexagon triple system is the set of all \(n \equiv 1\) or \(9 \pmod{18}\) and that if \((X,H)\) is a hexagon triple system of order n then \(| H| = n(n-1)/18\). Clearly the inside triples of a hexagon triple system cannot form a triple system, not even enough triples. Consequently the problem that remains to be addressed is that what is the largest Steiner triple system that can be embedded in the partial triple system consisting of the inside triples of a hexagon triple system. Alternatively the question may be posed as: Let \((X,H)\) be a hexagon triple system of order \(n\) and let the collection of inside triples as \(P\). Then \((X,P)\) is a partial triple system of order \(n\). Then the triple system \((S,T)\) is embedded in \((X,H)\) provided \(S\) contained in \(X\) and \(T\) contained in \(P\), and \(| S |\) to the largest possible one. Thus it is shown that any Steiner triple system of order \(n\) can be embedded in the inside triples of a hexagon triple system of order approximately \(3n\).