b
Let $\mathrm{I}_{1}$ be the current throug $5 \,\Omega$ resistance, $\mathrm{I}_{2}$ through $(6+9) \,\Omega$ resistance. Then as per question,
$\mathrm{I}_{1}^{2} \times 5=20$ or, $\mathrm{I}_{1}=2\, \mathrm{A}$
Potential difference across $C$ and $D=2 \times 5$ $=10 \,\mathrm{V}$
Current $I_{2}=\frac{10}{6+9}=\frac{2}{3}\, A$
Heat produced per second in $2\,\Omega$
$=\mathrm{I}^{2} \mathrm{R}\left(\frac{8}{3}\right)^{2} \times 2=14.2\,\mathrm{cal} / \mathrm{s}$