A coil of 45 mH is joined to a capacitor of 0.80 µF…

Step 1. The product. $LC = 45 \times 10^{-3}\ \text{H} \times 0.80 \times 10^{-6}\ \text{F} = 3.6 \times 10^{-8}\ \text{s}^2$. So $\sqrt{LC} = 1.90 \times 10^{-4}$ s.

Step 2. The angular frequency. $\omega_0 = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{1.90 \times 10^{-4}\ \text{s}} = 5270$ rad/s.

Step 3. The frequency. $f = \dfrac{\omega_0}{2\pi} = \dfrac{5270\ \text{rad/s}}{2\pi} = 839$ Hz.

Step 4. The period. $T = \dfrac{1}{f} = \dfrac{1}{839\ \text{Hz}} = 1.19\ \text{ms}$.