The power supplied by battery shown in the diagram is (ammeter is ideal)

Let I is current through battery 
Let x be current through 3$\mathrm{\Omega }$ resistance 

$\frac{\mathrm{x}\times 3}{15}=0.3\Rightarrow \mathrm{x}=1.5\mathrm{A}$

$\therefore \mathrm{I}=1.8\mathrm{A}$

$\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{Total} \mathrm{resistance} \mathrm{of} \mathrm{the} \mathrm{circuit}=$

$\frac{\left(9+6\right)\times 3}{\left(9+6\right)+3}+2=4.5\mathrm{\Omega }$

$\mathrm{P}={\mathrm{I}}^2\mathrm{R}=14.6\mathrm{W}$