Answer
${\bf 10.1}\;\rm mL$
Work Step by Step
We know, for $RLC$-circuits, that the resonance frequency occurs when
$$X_L=X_C$$
$$2\pi f L=\dfrac{1}{2\pi f C}$$
Hence,
$$ f^2 =\dfrac{1}{4\pi^2 L C}$$
Solving for $L$,
$$ L =\dfrac{1}{4\pi^2 f^2 C}$$
Plug the known;
$$ L =\dfrac{1}{4\pi^2 (1000)^2 (2.5\times 10^{-6})}$$
$$L=\color{red}{\bf 10.1}\;\rm mL$$
