A connected Graph has Euler Circuit $\iff$ all of its vertices have even degree
A connected Graph has Euler Path $\iff$ exactly 2 of its vertices have odd degree
(a) k-regular graph where k is even number.
a k-regular graph need not be connected always. Example : The given below graph is a $2$ regular graph is not a Euler graph. This is so because there is no single walk which covers all edges.
(b) the complete graph of $90$ vertices
In such a graph every vertex will have an odd degree = 89, Hence it cannot have a Euler path/Circuit.
(c) to get degree of all vertices of the complement of cycle on 25 vertices we need to subtract the degree of a complete graph of 25 vertices with degree of vertices in the original given graph i.e. cycle on 25 vertices.
Degree of complement $= 24 - 2 = 22$. Since, every degree is Even, and it is connected also, therefore Graph has a Euler Cycle.
It is connected because, there is a theorem which says, "$G$ be a graph with $n$ vertices and if every vertex has a degree of at least $\frac{n-1}{2}$ then $G$ is connected." [check this]
Here Degree of each vertex is $22$, which is of course greater that $\frac{25-1}{2}\left (=12 \right )$.
answer = Option C