How many coulombs of electricity are consumed when a

 100 mA current is passed through a solution of $A{\mathrm{gNO}}_3$ 

 for half an hour during an electrolysis experiment? 

      $1\mathrm{mA}={10}^{-3}\text{ ampere }$

$\therefore$       $100\mathrm{mA}\equiv 100\times {10}^{-3}\mathrm{A}$

             Time  $=1/2\mathrm{hr}=1/2\times 60\times 60=1800$ second

$\therefore$       Quantity of electricity to be passed 

               = Current in amperes x time in  second

 $\begin{array}{r}=100\times {10}^{-3}\mathrm{A}\times 1800\mathrm{s}=180\mathrm{As}\\ =180\mathrm{C}\end{array}$       

So, the correct answer is (b).