A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant 'b', the correct equivalence would be :

For a spring-mass damped oscillator

$\frac{d^{2}x}{d{t}^2}+\left(\frac{2x}{2\sqrt{mk}}\sqrt{\frac{k}{m}}\right)\frac{dx}{dt}+{\left(\sqrt{\frac{k}{m}}\right)}^{2}x=0$

$\Rightarrow \boxed{\frac{d^{2}x}{d{t}^2}+2\tau w_0\frac{dx}{dt}+w_0^{2}x=0}$

$\tau =\frac{c}{2\sqrt{mk}}\Rightarrow$ damping ratio $c=b\Rightarrow$ damping constant

$W_0=\sqrt{\frac{k}{m}}\Rightarrow$ angular frequency

$\Rightarrow \boxed{\frac{d^{2}x}{d{t}^2}+(\frac{b}{m})\frac{dx}{dt}+(\frac{k}{m})x=0}$..........(1)

For LCR circuit

use KVL:-

$V-L\frac{dI}{dt}-IR-\frac{q}{c}=0$

$\Rightarrow \boxed{\frac{d^{2}q}{d{t}^2}+\frac{R}{L}\frac{dq}{dt}+(\frac{1}{LC})q=\frac{V}{L}}$...........(2)     $I=\frac{dq}{dt}$

By comparing (1) and (2):=

$L\to m,R\to b,LC\to \frac{m}{k}$

                      $\Rightarrow C\to \frac{1}{k}$

option (D) is correct.