The correct answer is $\boxed{\left( {\frac{{\text{x}}}{{\sqrt 3 }},\,\frac{{\text{y}}}{{\sqrt 3 }},\,\frac{{\text{z}}}{{\sqrt 3 }}} \right)}$.
The unit normal vector at a point on the surface of a sphere is a vector that is perpendicular to the surface at that point and has a magnitude of 1. The vector $\left( {{\text{x, y, z}}} \right)$ is not perpendicular to the surface of the sphere at any point, so it is not the unit normal vector. The vector $\left( {\frac{1}{{\sqrt 3 }},\,\frac{1}{{\sqrt 3 }},\,\frac{1}{{\sqrt 3 }}} \right)$ is a unit vector, but it is not perpendicular to the surface of the sphere at any point, so it is not the unit normal vector. The vector $\left( {\frac{{\text{x}}}{{\sqrt 2 }},\,\frac{{\text{y}}}{{\sqrt 2 }},\,\frac{{\text{z}}}{{\sqrt 2 }}} \right)$ is not a unit vector, so it is not the unit normal vector.
The unit normal vector at a point on the surface of a sphere can be calculated using the following formula:
$$\hat{\mathbf{n}} = \frac{\mathbf{r}}{\|\mathbf{r}\|}$$
where $\mathbf{r}$ is the vector from the origin to the point on the surface of the sphere.
In this case, the point on the surface of the sphere is $(x, y, z)$, so the unit normal vector is:
$$\hat{\mathbf{n}} = \frac{\left( {x, y, z} \right)}{\sqrt{{x^2} + {y^2} + {z^2}}} = \left( {\frac{{\text{x}}}{{\sqrt 3 }},\,\frac{{\text{y}}}{{\sqrt 3 }},\,\frac{{\text{z}}}{{\sqrt 3 }}} \right)$$