The Handshaking theorem states that,
$\displaystyle{\sum_{v \in V}{deg (v_i)} = 2 \times \mid E\mid}$
Which means, $\text{Each edge contributes twice to the sum of the degrees of all vertices.}$
$\text{Sum of degree of vertices = 2 } \times Edges$
Now, the graph has $35$ Edges.
∴ $2 \times Edges = 2 \times 35 = 70$ is the Maximum Sum of degree of the vertices.
We don't know the number of vertices.
So, we assumed the total number of vertices will be $n$
Condition given is "All vertices are of degree at least $3$"
Sum of the degree of vertices = $3 \times n$ $\qquad \left[ \text{∵ All vertices are of at least degree 3} \right ]$
∴ Minimum sum of degree of vertices will be $3n$ which will be less than or equal to $70$ because $70$ is the Maximum Sum of degree of vertices.
∴ $3\times n \leq 2\times(35)$
Or, $n \leq \dfrac{70}{3}$
Or, $n \leq 23.33$
∴ $\color{Maroon}{\text{Largest possible number of vertices in the graph must be }} \color{Green}{23}$ $\color{teal}{\text{ as we can't take the fractions & will take only the whole number.}}$