Calculating base dissociation constant given the molarity and pH
$\ce{C6H5NH2}$ (aniline) is a weak base, and therefore, in its aqueous solution, it takes a proton from water and forms $\ce{C6H5NH3^+}$ and $\ce{OH-}$, which are correct as you have written.

Suppose the extent of protonation is $\alpha$. After taking up the proton, $[\ce{C6H5NH2}] = c (1 - \alpha)$ and $[\ce{C6H5NH3^+}]= c \alpha = [\ce{OH-}]$.

So, we can write the following, from the above mentioned reaction:

$$
K_{\mathrm{b}} = \frac{[\ce{C6H5NH3^+}][\ce{OH-}]}{[\ce{C6H5NH2}]} = \frac{c^{2}\alpha^{2}}{c(1 - \alpha)} \approx c \alpha^{2}
$$

As the degree of protonation is very less than 1 ($\alpha &lt;&lt; 1$), we get $\alpha = \sqrt{K_{\mathrm{b}} / c}$. Now,

\begin{align*}
\mathrm{p}\ce{OH} &amp;= -\log[\ce{OH-}] \\
&amp; = -\log(c\alpha)=-\log(c\sqrt{K_{\mathrm{b}}/ c})=-\log(\sqrt{cK_{\mathrm{b}}}) \\
&amp;= \frac{\mathrm{p}K_{\mathrm{b}}}{2} - \frac{\log c}{2}
\end{align*}

(as $-\log K_{\mathrm{b}} = \mathrm{p}K_{\mathrm{b}}$).

In the question, $\mathrm{pH} = 8.8$, so, $\mathrm{pOH} = 14 - 8.8 = 5.2$, and $c = 0.1$. Putting these values in the equation we will get

$$
\mathrm{p}K_{\mathrm{b}} = 2.(5.2 + \frac{\log(0.1)}{2}) = 9.4 \implies K_{\mathrm{b}} = 10^{-9.4} = 3.98 \times 10^{-10}
$$

Thus we get the $K_{\mathrm{b}}$ of aniline as $3.98 \times 10^{-10}$.

Note that, if you have any aqueous solution consisting of only weak base, you can apply the direct formula as $\mathrm{pOH} = \frac{\mathrm{p}K_{\mathrm{b}}-\log c}{2}$, and easily get the $K_{\mathrm{b}}$.