Solution:

For a uniformly charged ring:

1. Electric potential on the axis:
\[
V = \frac{kQ}{\sqrt{R^2 + x^2}},
\]
where \( R = \sqrt{5} \) and \( x \) is the distance from the center.

2. Electric field on the axis:
\[
E = \frac{kQx}{(R^2 + x^2)^{3/2}}.
\]

3. Given:
\( V = 6 \, \text{V} \), \( E = 1 \, \text{V/m} \).

4. Divide \( E \) by \( V \) to eliminate \( kQ \):
\[
\frac{E}{V} = \frac{x}{R^2 + x^2} \quad \Rightarrow \quad \frac{1}{6} = \frac{x}{5 + x^2}.
\]

5. Simplify:
\[
5 + x^2 = 6x \quad \Rightarrow \quad x^2 - 6x + 5 = 0.
\]

6. Solve quadratic:
\[
(x - 5)(x - 1) = 0 \quad \Rightarrow \quad x = 5 \, \text{or} \, x = 1.
\]

The smallest value of \( x \) is 1 m.