${\mathrm{pK}}_{\mathrm{b}}$ of aq.${\mathrm{NH}}_3$ is $4.74$ hence $\mathrm{pH}$ of $0.01\mathrm{M}$ ${\mathrm{NH}}_3$ solution is

The pH of the solution can be calculated as:

$pH=14-pOH$

The pOH can be calculated as:

 pOH=$\frac{1}{2}{p}^{K_b}-\frac{1}{2}\log \left[C\right]$

 pOH=$\frac{1}{2}(4.74)-\frac{1}{2}\log [0.01]$

 pOH=$\frac{1}{2}(4.74)+\frac{1}{2}\left(2\right)$

 pOH= 2.37 + 1 = 3.37 

Now putting the value in pH equation:

$pH=14-pOH$

$pH=14-3.37=10.63$

Therefore, the correct option is (B).