How to get the LC from a transfer function while given a crossover frequency (low pass RLC filter)
It's not enough to have the corner frequency as the only requirement (and the R), because, as you can see from your transfer function, the quality factor is dependent on C. Thus, L=1m C=1u will not be the same as L=1u C=1m. Once you have your quality factor, Q, or the damping, ζ, you can use whatever way feels easier for you to solve for L and C. For example:
$$
\dfrac{\dfrac{1}{LC}}{s^2+\dfrac{1}{RC}s+\dfrac{1}{LC}}=\dfrac{\omega^2}{s^2+\dfrac{\omega}{Q}s+\omega^2} \tag{1}
$$
From here, a system of two equations is formed:
$$\left\{\begin{aligned}
\omega^2&amp;=\dfrac{1}{LC} \\
Q&amp;=\omega RC \\
\end{aligned}\right. \tag{2}$$
$$\left\{\begin{aligned}
L&amp;=\dfrac{R}{\omega Q} \\
C&amp;=\dfrac{Q}{\omega R}
\end{aligned}\right. \tag{3}$$
Suppose you want a Bessel response, unscaled. Then Q=1/√3, and for R=50 Ω, f=1 kHz =&gt; L=13.78 mH and C=1.84 μF. It's useless to talk about a stopband frequency since you have a fixed 2nd order filter which has a fixed slope of -40 dB/dec.