Let the min-degree of $G$ is $x$. then $G$ has at least $\left[|v|\times \frac{x}{2}\right]$ edges.
$\left[|v|\times \frac{x}{2}\right]\leq \left[\left(3\times |v|\right) -6\right]$
for $x=6$, we get $0 \leq {-6}$, Therefore, min degree of $G$ cannot be $6$.
Correct answer is (D).
$\text{An alternative approach,}$
Let the min_degree of a graph is $\text{'x'}$ , then
$x\leq \left(\frac{2e}{n}\right)$,
given , $e\leq \left(3n - 6\right)$ $\text{\{it will be planner graph\}}$
put the value of $e$ ,then min_degree will be ,
$x\leq \frac{(2(3n-6))}{n}$
$x\leq \frac{\left(6n - 12\right)}{n}$
$x\leq \left( \frac{6n}{n} - \frac{12}{n} \right)$
$x\leq \left(6 - \frac{12}{n}\right)$ ,
when number of vertices is more , then value of
$\left(\frac{12}{n}\right)$ will be less , $\left(\frac{12}{n} = 0.000001 \text{assume}\right)$ ,
then min_degree will be ,
$x\leq \left(6 - 0.000001\right)$
$x\leq 5.999999$ , max value
$x\leq \text{floor value}\left(5.9999999\ldots \right)$
$x = 5$ , maximum value of min_degree of defined graph (i.e. planner graph)