$\text{The resistivity of a potentiometer wire is}10{\mathrm{-}}^7\Omega \text{m and its cross-section area is}\mathrm{ 10}{\mathrm{-}}^6{m}^2.$

$\text{When a current of}\mathrm{ 0.1}\text{A flows through the wire, its potential gradient is:}$

The correct option is:
${10}^{-2}\text{V/m}$

Step-by-step Explanation:

1. Identify the given values:
Let us note the given parameters from the problem statement:
• Resistivity of the potentiometer wire ($\rho$): ${10}^{-7}\text{Ω m}$
• Cross-sectional area of the wire ($A$): ${10}^{-6}\text{m}{s}^2$ (which is $m^2$)
• Current flowing through the wire ($I$): $0.1\text{A}={10}^{-1}\text{A}$

2. Understand the formula for potential gradient:
The potential gradient ($k$) is defined as the potential drop per unit length of the wire:
$$k=\frac{V}{L}$$
By Ohm's law, the potential difference across a length $L$ of the wire is:
$$V=I·R$$
We also know that the electrical resistance ($R$) of the wire of length $L$ is given by:
$$R=\rho ·\frac{L}{A}$$
Substituting the expression for $R$ back into the voltage equation:
$$V=I·\left(\rho ·\frac{L}{A}\right)$$
Substituting this expression for $V$ back into the potential gradient formula gives:
$$k=\frac{I·\rho ·L}{A·L}=\frac{I·\rho }{A}$$

3. Calculate the potential gradient:
Now, substitute the given values into the simplified equation:
$$k=\frac{\left({10}^{-1}\text{A}\right)·\left({10}^{-7}\text{Ω m}\right)}{{10}^{-6}\text{m}{s}^2}$$
Combine the powers of 10 in the numerator:
$${10}^{-1}·{10}^{-7}={10}^{-8}$$
Now, divide by the denominator:
$$k=\frac{{10}^{-8}}{{10}^{-6}}={10}^{-8-\left(-6\right)}={10}^{-2}\text{V/m}$$

Therefore, the potential gradient of the potentiometer wire is ${10}^{-2}\text{V/m}$.