How do I find the power of this circuit?
The voltage across the resistor is

$$V_{R1} = 2\mathrm A \cdot 10 \Omega  = 20V$$

with the more positive terminal on the left (since the current from the current source enters the left-most resistor terminal.

Thus, the voltage across the current source (being careful to choose the reference direction such that the current enters the positive labelled terminal per the passive sign convention) is, by KVL:

$$V_{I1} = (V_{V1} - V_{R1})\mathrm V = (14 - 20)\mathrm V = -6\mathrm V$$

The power associated with the current source is then

$$P_{I1} = V_{I1} \cdot I_{I1} = -6\mathrm V \cdot 2\mathrm A = -12\mathrm W$$

By the passive sign convention, when the power is negative, the circuit element is supplying power to the circuit thus, the current source supplies \$12 \mathrm W\$ of power to the circuit.

As a check, the sum of the powers should be zero.

The power delivered to the resistor is:

$$P_{R1} = 20 \mathrm V \cdot 2 \mathrm A = 40 \mathrm W$$

The power associated with the voltage source is

$$P_{V1} = -14 \mathrm V \cdot 2 \mathrm A = -28 \mathrm W$$

The sum of the powers is then

$$P_{I1} + P_{R1} + P_{V_1} = (-12 +40 -28) \mathrm W = 0 \mathrm W$$

as required.

Note that another approach would be to calculate the resistor and voltage source powers using the series current and write, using the fact that the sum of powers is zero:

$$P_{I1} = -(P_{R1} + P_{V1}) =  -\left((2\mathrm A)^2 \cdot 10 \Omega - 14\mathrm V \cdot 2\mathrm A)\right) = -12\mathrm W$$

as before.