When a potential difference V is applied across a wire of...

For first case:

$\dfrac{{V}^{2}}{R} = W$     ...(i)

Now, equivalent resistance of the combination will be $\dfrac{\dfrac{R}{2} \times \dfrac{R}{2}}{\dfrac{R}{2} + \dfrac{R}{2}} = \dfrac{R}{4}$

For second case:

${W}^{'} = \dfrac{{V}^{2}}{\dfrac{R}{4}}$    ...(ii)

From (i) & (ii), we get
${W}^{'} = 4W$