I hope I wrote the question well since it is my attempt at translating from the book. If it isnt clear enough, The question states that in every graph as described in the title a simple cycle of length 4 exists.
Sorry for my english and thanks in advance for the help.
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Clearly, there is a Hamiltonian cycle, say $abc....a$. [since, $\delta(G) \geq n/2$] (You don't need this though, but was the first thought that came to me.)
If $G$ is a complete graph you are done, since $G$ has $k$-cycles for $k=3,4,...,100$. Else $G$ has a triplet $abc$ which is a path.
Now consider this path $abc$, since $deg(a)$ and $deg(c)$ is greater than 50, we need atleast 49 vertices (except $b$) each to be neighbours of $a$ and for $c$, so a total of $98$ vertices. But we have already used up $3$ vertices, hence are left with 97 more vertices. So there exists a vertex $d \neq b$ which is common in $nbd(a)$ and $nbd(c)$.
So $abcda$ is a four cycle. Infact for every induced path $abc$ we have a 4-cycle.
We may assume that the graph is not complete. Choose two nonadjacent vertices $u$ and $v$. Now $N(u)$ and $N(v)$ are $50$-element subsets of the $98$-element set $V(G)\setminus\{u,v\}$, so $$|N(u)\cap N(v)|=|N(u)|+|N(v)|-|N(u)\cup N(v)|\ge50+50-98=2.$$ So $u$ and $v$ and their two common neighbors form a subgraph $K_{2,2}=C_4$ in $G$.