Current measured by the ammeter A in the reported circuit when no current flows through 10 Ω resistance…

This is a Wheatstone circuit. For the current to be zero in $10 \mathrm{\Omega }$ the Wheatstone circuit should be balanced. Therefore,

$\frac{R}{3}=\frac{4}{6}\Rightarrow R=2 \mathrm{\Omega }$

Now, the equivalent resistance across the battery will be, 

$R_{eq}=\frac{\left(2+4\right)\times \left(3+6\right)}{\left(2+4\right)+\left(3+6\right)}=3.6 \mathrm{\Omega }$

Therefore, the current measured by the ammeter is $i=\frac{V}{R}=\frac{36}{3.6}=10 \mathrm{A}$.