= Solution
Write $\gamma_{ij}={}^{(3)}g_{ij}$, $\gamma=\det\gamma_{ij}$ and $D_i$ for the <spatial covariant derivative>. With the negative-shift convention, the <extrinsic curvature of a spatial hypersurface> is $K_{ij}=-(\dot\gamma_{ij}+D_iN_j+D_jN_i)/(2N)$. Contracting with $\gamma^{ij}$ and using metric compatibility gives
$$
K=-\frac1{2N}\left(\gamma^{ij}\dot\gamma_{ij}+2D_iN^i\right).
$$
The logarithmic <determinant> identity makes $\gamma^{ij}\dot\gamma_{ij}=\dot\gamma/\gamma$, hence
$$
\boxed{K=-\frac1{2N}\left(\frac{\dot\gamma}{\gamma}+2D_iN^i\right).}
$$
This is also the fractional volume expansion, with a minus sign from the extrinsic-curvature convention.
Define $a=\gamma^{1/6}$ so that $\sqrt\gamma=a^3$. The useful unit-determinant conformal decomposition is $\gamma_{ij}=a^2\widetilde\gamma_{ij}$ with $\det\widetilde\gamma=1$, equivalently $\widetilde\gamma_{ij}=a^{-2}\gamma_{ij}$. Multiplying by $a^2$ instead would be a different conformal rescaling, not this unit-determinant one; no conformal decomposition is needed for the trace identity itself.
With zero <shift vector>, $\dot\gamma/\gamma=6\dot a/a$ and the <local volume Hubble parameter> is
$$
\boxed{H=-\frac K3=\frac1{Na}\dot a.}
$$
Along a normal trajectory proper time satisfies $ds=Ndt$, so $H=a^{-1}da/ds$ is the local physical volume-expansion rate. The printed equality with $\dot a/a$ requires the proper-time gauge $N=1$; zero shift alone does not set the lapse to unity. In a homogeneous <Friedmann-Lemaître-Robertson-Walker metric> it reduces to the ordinary <Hubble parameter>, while in an inhomogeneous spacetime it can vary from one normal trajectory to another.
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