The sum of the squares of two consecutive even natural numbers is 3700. The sum of the numbers is:

The correct option is 86.

Let the two consecutive even natural numbers be represented as $x$ and $x+2$, where $x$ is an even natural number.
According to the problem, the sum of the squares of these two consecutive even numbers is 3700.
We can write this condition mathematically as:

$x^2+{(x+2)}^2=3700$

We expand the second term using the algebraic identity ${(a+b)}^2=a^2+2ab+b^2$:

$x^2+(x^2+4x+4)=3700$

Combine the like terms:

$2x^2+4x+4=3700$

Subtract 3700 from both sides to form a quadratic equation:

$2x^2+4x-3696=0$

Divide the entire equation by 2 to simplify it:

$x^2+2x-1848=0$

Now, we solve this quadratic equation using factorization. We need to find two numbers that multiply to -1848 and add up to 2. These numbers are 44 and -42.
Rewriting the middle term:

$x^2+44x-42x-1848=0$

Group the terms to factorize:

$x(x+44)-42(x+44)=0$

$(x-42)(x+44)=0$

This gives two possible values for $x$:
$x=42$ or $x=-44$.
Since the question specifies that they are natural numbers, $x$ must be positive. Therefore, we discard $x=-44$.
So, the first even natural number is:

$x=42$

The second consecutive even natural number is:

$x+2=44$

The question asks for the sum of these two numbers:
Sum = $42+44=86$.