Answer
$P= \frac{9}{32}W$
Work Step by Step
let $ I_1$ be the current through $12V $ battery and $ I_2$ be current through $15V $ battery. let their directions are clockwise. so current through $2\Omega $ is $I_1-I_2$ downward
applying kirchoffs laws to each loop
first loop
$ 12V - 4I_1 - 2*(I_1-I_2) =0$
2nd loop
$15V - 4I_2 +2(I_1-I_2) =0$
subtracting first from second equation we get
$3V +8(I_1-I_2) =0$
$ I_1 -I_2 = \frac {3}{8}A$
thus power dissipated is
$ P = I^2R = (\frac{3}{8})^2 *2 W$
$P= \frac{9}{32}W$
