= Solution
For a <Keplerian accretion disk>, $\Omega=(GM_{\rm BH}/R^3)^{1/2}$, so
$$
F(R)=\frac{9GM_{\rm BH}}{4R^3}\nu\Sigma
=\frac{3GM_{\rm BH}\dot m}{4\pi R^3}
\left[1-\left(\frac{R_{\rm ISCO}}R\right)^{1/2}\right].
$$
Here $F$ is the dissipation summed over the two faces. The total luminosity is
$$
L=\int_{R_{\rm ISCO}}^\infty2\pi R F(R)\,dR
=\boxed{\frac{GM_{\rm BH}\dot m}{2R_{\rm ISCO}}}.
$$
This is the <Newtonian gravitational potential energy>[binding energy] per unit mass of a circular orbit at the inner edge, multiplied by the accretion rate. The <zero-torque inner boundary condition> ensures that no additional mechanical work enters from smaller radii.
The luminosity emitted outside radius $R$ is
$$
L(>R)=\frac{3GM_{\rm BH}\dot m}{2R}
\left[1-\frac23\left(\frac{R_{\rm ISCO}}R\right)^{1/2}\right].
$$
Writing $y=(R_{\rm ISCO}/R)^{1/2}$, the condition $L(>R)=L/2$ becomes $3y^2-2y^3=1/2$. Its physical root is $y=1/2$, and hence
$$
\boxed{R_{1/2}=4R_{\rm ISCO}}.
$$
Half of a thin disk's luminosity therefore comes from only the innermost factor four in radius, so strong-gravity effects and the inner-boundary condition strongly influence its observed spectrum.
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