Answer
11 volts.
Work Step by Step
First find the current in the entire circuit, from the given resistance and the power dissipated in it.
$$P=I^2R$$
$$I=\sqrt{\frac{P_{33}}{R_{33}}}=\sqrt{\frac{0.80W}{ 33\Omega}}=0.1557A$$
The circuit consists of a $68\Omega$ resistor in parallel with a $85\Omega$ resistor, and the combination is in series with the $33\Omega$ resistor.
$$R_{eq}=(\frac{1}{68\Omega }+\frac{1}{85\Omega })^{-1}+33\Omega=70.78\Omega$$
From Ohm’s law, V=IR, the battery voltage V can be found by multiplying the circuit current by the circuit’s equivalent resistance.
$$V=IR_{eq}=(0.1557A)( 70.78\Omega)=11V$$