Bandpass filters usually employ both inductors and capictors. The inductor cuts off the high frequencies, while the capacitor cuts off the low frequencies, ideal for a bandpass filter.
Whoops, I just noticed that the the answer is given in the figure. I can see they did not use a capacitor! I managed to derive the transfer function for this circuit and found the following, [tex]T(jw) = \frac{-z}{(jwL + z)(jwL + R)}[/tex] Now we simply need to select our Z such that the requirement for resonant frequency is obatined. If we select, [tex]z = R_{z}[/tex] We rearrange the transfer function, [tex]\frac{-\frac{1}{R}}{(1 + \frac{jwL}{R_{z}})(1 + \frac{jwL}{R})}[/tex] This becomes, [tex]\frac{-\frac{1}{R}}{(1 + \frac{jw}{w_{L}})(1 + \frac{jw}{w_{H}})}[/tex] Now we know what wL and wH are we can solve for fL and fH since, [tex]f = 2\pi w[/tex] Then our desired resonant frequency, [tex]f_{r} = \sqrt{f_{L} \cdot f_{H}}[/tex] Simply solve for, [tex]R_{z}[/tex]. You should find that, [tex]R_{z} = \frac{100\pi^{2}L^{2}}{R}[/tex]