$40$ ml of a hydrocarbon undergoes combustion in $260$ml of oxygen and gives $160$ml of $C{O}_2$. If all volumes are measured under similar conditions, the formula of the Hydrocarbon is 

We know that the combustion reaction is represented as:

$C_x{H}_y+(x+y/4)O_2------> xC{O}_2+y/2H_{2}O$

$40$ml         $260$ml                                          $160$ml

For $1$ml of $C_x{H}_y$:

$40/40$             $260/40$                             $160/40$

      $1$                    $6.5$                                           $4$

X = $4$ as $1$mol of hydrocarbon is producing $4$ mol of carbon dioxide. $x+y/4=6.5$

y =$10$

So, formula of hydrocarbon is $C_4{H}_{10}$ that is butane.

Therefore, Option (D) is correct.