A constant voltage is applied on a uniform wire, then the heat is produced.  The heat so produced will be doubled, if     

$H =\frac{V^2}{R}$   ------------------------- (1)

$Now we have, R = \frac{\rho L}{A}$
$Therefore, H = \frac{V^2 A}{\rho L}$
$A = \pi r^2$ here 'r' is the radius.

Hence, if the radius is doubled then, new area would be 4A 
$R = \frac{\rho \left(2L\right)}{4A} = \frac{\rho L}{2A}$; The resistance is halved

From equation no. (1), the heat energy would be doubled.