= Solution
A generalized <conformal metric> is locally $ds_\rho=\rho(z)|dz|$, where $\rho$ is nonnegative and measurable and transforms as a length density under a <holomorphic> change of coordinate. Its area is $A_\rho=\int_X\rho^2\,dx\,dy$. Use the usual convention of using <locally rectifiable paths> for a <path family> $\Gamma$, and put $L_\rho(\Gamma)=\inf_{\gamma\in\Gamma}\int_\gamma\rho|dz|$. Then
$$
\boxed{\lambda(\Gamma,X)=\sup_{\rho:\ 0<A_\rho<\infty}
\frac{L_\rho(\Gamma)^2}{A_\rho}.}
$$
This is <extremal length>. Zeros and isolated singularities of an admissible density are allowed; requiring a <smooth> strictly positive <Riemannian metric> would unnecessarily restrict the definition. Line integrals have their extended nonnegative values; if an arbitrary family is supplied, use its members that are <locally rectifiable paths>. An empty <path family> has infinite infimal length, whereas a family containing a constant path has <extremal length> zero.
Both numerator and denominator scale quadratically when $\rho$ is multiplied by a positive constant. The coordinate transformation of the area element makes the quotient unchanged under <conformal equivalence>.
The normalization relevant later is worth deriving. On the <conformal cylinder> $(\mathbb R/\mathbb Z)\times(0,M)$, let $\Gamma$ contain the loops going once around it. For the horizontal loop at height $t$, <Cauchy-Schwarz inequality> gives
$$
\left(\int_0^1\rho(x,t)\,dx\right)^2\leq\int_0^1\rho(x,t)^2\,dx.
$$
Integrating in $t$ shows $L_\rho(\Gamma)^2\leq A_\rho/M$. The constant density achieves equality, since every winding-one loop has Euclidean length at least one. Hence
$$
\boxed{\lambda(\Gamma,\text{cylinder})=\frac1M.}
$$
Here $M$ is the height divided by circumference, the <conformal modulus of an annulus>; the reciprocal is used for the family joining its boundary components.
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