In this case, $V = 22.0 \,\text{V}$ and $I = 3.00 \,\text{A}$, so the power is:
$$P_{battery} = (22.0 \,\text{V})(3.00 \,\text{A}) = 66.0 \,\text{W}$$
(b) The power being delivered to the resistance of the coil is given by $P_{resistance} = I^2R$. In this case,
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