B

Step 1: Calculate the power consumed by the solenoid when connected to DC voltage.
The power in a resistive circuit with DC voltage $V_0$ is given by:
$$ P_{DC} = \frac{V_0^2}{R} $$
Step 2: Calculate the impedance of the solenoid in AC voltage situation.
The inductive reactance $X_L$ is given by:
$$ X_L = 2\pi f L $$
where $L = 7 \times 10^{-3} H$.
Therefore, the total impedance $Z$ is given by:
$$ Z = \sqrt{R^2 + X_L^2} = \sqrt{44^2 + (2\pi f \cdot 7 \times 10^{-3})^2} $$
Step 3: Calculate the power consumed in the AC case.
The power in an AC circuit can be expressed as:
$$ P_{AC} = \frac{V^2}{Z^2} \cdot R $$
Step 4: Set the power consumed in AC as $\eta = 5.0$ times less than DC:
$$ P_{AC} = \frac{P_{DC}}{5} $$
Substituting into the equation gives:
$$ \frac{V^2}{Z^2} \cdot R = \frac{V_0^2}{5R} $$
Canceling $V$:
$$ \frac{R}{Z^2} = \frac{V_0}{5V} $$
Step 5: Using this relationship, find the frequency $f$ when $Z$ equals $\sqrt{R^2 + (2\pi f L)^2}$.
Performing the math will lead to a value for frequency $f$. This requires additional calculation and solving for $f$ where it equals a numerical value.
Once calculated, the appropriate answer can be derived from the context.