Your "answer" isn't even close to being correct. Do you know how two (or more) resistors in series can be replaced with a single equivalent resistance? Do you know how two (or more) resistors in parallel can be replaced with a single equivalent resistance? You must reduce your circuit to a single equivalent resistance connected to the 8V battery by combining series and parallel equivalent resistances. Start with the series-connected 15Ω and 7Ω resistors, which are in parallel with the 22Ω resistor. Work from right to left until you have a single resistance connected to the 8V battery.
For two resistances, R1 and R2, in series the equivalent resistance Rseries = R1 + R2. For two resistances, R3 and R4, in parallel the equivalent resistance Rparallel = (R3)(R4) / (R3+R4). Another way to write this is 1/Rparallel = 1/R3 + 1/R4. Or in words, the reciprocal of two resistances in parallel is the sum of their reciprocal resistances.
So, if you draw your sideways circuit out so you can see it properly, and then proceed to reduce the circuit to a single equivalent resistance connected across the 8V battery, you can calculate the current this equivalent resistance draws from the 8V battery. This will be the same current that is drawn through the 6Ω resistor. Why? Because all the current drawn from the battery passes through the 6Ω resistor. The square of this current multiplied by 6Ω will tell you how much power is dissipated in the 6Ω resistor, which was the original question that was asked.
