Reasoning about equal current on both sides of resonance.
The current in a series RLC circuit depends on capacitance only through the net reactance $X = X_L - X_C$. Current is maximum at resonance, where $X = 0$, and falls off symmetrically as $X$ moves away from zero in either direction. So two different capacitor settings give the same current only if they sit symmetrically about resonance: one with a net inductive reactance $+x$ and the other with net capacitive reactance $-x$.
That symmetry condition is \[X_L - \frac{1}{2\pi f C_1} = -\left(X_L - \frac{1}{2\pi f C_2}\right).\]
Adding $X_L$ terms together: \[2X_L = \frac{1}{2\pi f C_1} + \frac{1}{2\pi f C_2}.\]
With $X_L = 2\pi f L$: \[2\cdot 2\pi f L = \frac{1}{2\pi f}\left(\frac{1}{C_1} + \frac{1}{C_2}\right).\]
Combine the fractions using $\dfrac{1}{C_1} + \dfrac{1}{C_2} = \dfrac{C_1 + C_2}{C_1 C_2}$: \[4\pi f L = \frac{1}{2\pi f}\cdot\frac{C_1 + C_2}{C_1 C_2}.\]
Dividing both sides by $4\pi f$ gives \[L = \frac{1}{8\pi^2 f^2}\cdot\frac{C_1 + C_2}{C_1 C_2}.\]
Note that $R$ drops out completely, which is why the option containing $R$ cannot be correct. \[\boxed{L = \frac{1}{8\pi^2 f^2}\,\frac{C_1 + C_2}{C_1 C_2}}\]