Find the pH of a sodium chloride solution using the extended Debye-Huckel equation
Let's first look at ionic strength:

Ionic strength is a measure of the total ion concentration in solution, however, species with higher charge can also have elctrostatic interactions with other species.

$$\mu = \frac{1}{2} \sum_i C_iZ_i^2 $$

What is the strength of 0.02 M NaCl solution?

$$\mu = \frac{1}{2}([\ce{Na+}]z^2 + [\ce{Cl-}]z^2) = \frac{1}{2}([0.02](1)^2 + [0.02](-1)^2 ) = 0.02 M$$

With the ionic strength, we can now calculate the activity coefficients ($\gamma_i$) of ions in solution. The activity is merely the product of the concentration of each species and the activity coefficient. 
$$\textrm{activity} = a_i = \gamma_i[i]$$

Let's look at a typical equilibrium expression:

$$\ce{A + B &lt;=&gt; C + D} $$

$$K_{eq} = \frac{[C]\gamma_c[D]\gamma_d}{[A]\gamma_a[B]\gamma_b}$$

The Extended Debye-Huckel equation is used to find the effect of ionic strength on the activity coefficent. It is given as above:

$$\log_{\gamma \ce{H+}}=\frac{-.51(z_i^2)(\sqrt{\mu})}{1+(\alpha(\frac{\sqrt{\mu}}{305})}$$

where $\alpha$ is the size of the hydrated ion in pm.

The Extended Debye-Huckel equation is valid from $\mu = 0 \ \text{to} \ 0.1 \ \text{M}.$

To find pH, you must first find $\mu$ of the hydrogen ion. You can then use the Extended Debye-Huckel equation to find the $log_{\gamma \ce{H+}}$. Take the antilog.

$$a_{\ce{H+}} = \gamma_{\ce{H+}}[\ce{H+}] $$

Take the negative log and you will get pH.

$$ pH = -\log( a_{\ce{H+}} ) = -\log( \gamma_{\ce{H+}}[\ce{H+}] ) $$

In context with the problem:

$$\log_{\gamma_{\ce{H+}}} = \frac{-.51(1^2)(\sqrt{0.02})}{1+(900 (\frac{\sqrt{0.02}}{305})}$$

$$\log_{\gamma_{\ce{H+}}} = -0.0508886 $$

$$\gamma_{\ce{H+}} = 1.1243166 $$

$$\textrm{pH} = -\log a_{\ce{H+}} = -\log (\gamma_{\ce{H+}} \cdot 1.0 \cdot 10^{-7}) = 6.95 $$

Your error is when you take the antilog of the $\log_{\gamma \ce{H+}}$. You raised it to the $10^{\log_{\gamma \ce{H+}}}$ which is $0.89429197$. It is actually supposed to be $10^{-(-0.0508886)}$ which comes out to be $1.124316$.

For the 0.10 M of $\ce{NaCl}$ I calculated the pH to be $6.92$.