\text{The roots of the equation}\mathrm{ a}{x}^3-24x^2+188x-480=0\text{ are three consecutive even natural numbers. The value of}\mathrm{ a}\text{ is \_\_\_\_\_\_\_\_\_\_.}

The correct option is 1.

We are given a cubic equation:
$$a{x}^3-24x^2+188x-480=0$$

Let the roots of this cubic equation be three consecutive even natural numbers. We can denote these roots as:
$$\alpha =2k-2,\beta =2k,\gamma =2k+2$$ where $k$ is an integer such that $k\ge 2$ (so that the roots are positive even integers, i.e., natural numbers).

For a standard cubic equation of the form $a{x}^3+b{x}^2+cx+d=0$ with roots $\alpha$, $\beta$, and $\gamma$, the relations between the coefficients and the roots are given by Vieta's formulas:
1. Sum of the roots:
$$\alpha +\beta +\gamma =-\frac{b}{a}=\frac{24}{a}$$ 2. Sum of the product of roots taken two at a time:
$$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}=\frac{188}{a}$$ 3. Product of the roots:
$$\alpha \beta \gamma =-\frac{d}{a}=\frac{480}{a}$$

Substituting the expressions for the roots in terms of $k$ into the first relation:
$$(2k-2)+2k+(2k+2)=\frac{24}{a}$$ Simplifying the left-hand side:
$$6k=\frac{24}{a}\Rightarrow a=\frac{4}{k}$$

Now, substitute this relation $a=\frac{4}{k}$ into the third relation (product of the roots):
$$(2k-2)\left(2k\right)(2k+2)=\frac{480}{a}$$ Using the identity $(2k-2)(2k+2)=4k^2-4$, we have:
$$2k(4k^2-4)=\frac{480}{(4/k)}$$ $$8k(k^2-1)=120k$$

Since the roots are consecutive natural numbers, we know that $k>0$. We can safely divide both sides by $8k$:
$$k^2-1=15$$ $$k^2=16$$ Since $k$ must be a positive integer for the roots to be positive natural numbers, we take:
$$k=4$$

Substituting $k=4$ back into our equation for $a$:
$$a=\frac{4}{k}=\frac{4}{4}=1$$

Thus, the value of $a$ is $1$.