I always manage to confuse myself when it comes to concepts related to how light or heat is radiated, and in particular, when it comes to the basic formulae concerning Lambertian surfaces. It is time to work things out from the basics, from scratch.

First, the simplest case: A sphere with a surface of $A$, emitting light uniformly in all sky directions. If its power is $P$ (this would be measured, e.g., in watts), its radiant intensity is (with SI units in parenthesis):
\begin{align*} I=\dfrac{1}{4\pi}P~({\rm W}/{\rm sr}). \end{align*}
\begin{align*} J=\dfrac{P}{A}~({\rm W}/{\rm m}^2). \end{align*}
and its radiance is
\begin{align*} L=\dfrac{1}{4\pi}\dfrac{P}{A}=\dfrac{1}{4\pi}J~({\rm W}/{\rm m}^2/{\rm sr}). \end{align*}
(I am using the notation and terminology from Wikipedia.)

So let me move on to a surface element $dA$ in the source plane, with radiosity $J.$ Suppose that this surface element radiates equally in all hemispherical sky directions. In that case, its radiance is
\begin{align*} L=\dfrac{1}{2\pi}J~{\rm W}/{\rm m}^2/{\rm sr}. \end{align*}

The generic relationship between these quantities is as follows:
\begin{align*} P&=\iint J~dA,\\ P&=\iint I~dS,\\ I&=\iint L~dA,\\ J&=\iint L~dS,\\ P&=\iiiint L~dA~dS, \end{align*}
where $dA$ represents a surface element of the emitting surface, and $dS$ represents a sky element (in units of sr).

In the usual spherical coordinates, $dS=\sin\theta d\theta d\phi$, where $\theta$ is the angle between the sky direction and the surface normal and $\phi$ is the azimuthal angle.

Surface elements do not usually emit equally in all sky directions. A diffuse surface obeys Lambert's cosine law: its radiance is proportional to the cosine of the angle $\theta$ between the surface normal and the direction of emission, $L=\cos\theta L_0$, with $0\le\theta\le\pi/2$ and $L_0$ is the radiance in the normal direction. Therefore,
\begin{align*} J=\iint L~dS=L_0\int_0^{2\pi}d\phi\int_0^{\pi/2}d\theta\cos\theta\sin\theta=\pi L_0. \end{align*}
Therefore, the radiance in the normal direction is given in terms of the radiosity as
\begin{align*} L_0=\frac{1}{\pi}J. \end{align*}

Let me move on to a luminous sphere, characterized by spherical coordinates $\vartheta$, $\psi$, and viewed from the $\vartheta=0$ direction. The relationship between a surface element $d\Sigma$ on the sphere and the corresponding projected surface element $dA$ in the image plane is given by
\begin{align*} dA=\cos\vartheta d\Sigma. \end{align*}

The surface element $d\Sigma$ will emit light in accordance with Lambert's cosine law with radiance $L d\Sigma=L_0\cos\vartheta d\Sigma=L_0 dA$ in the viewing direction. In other words, the radiance of the projected disk is constant throughout the disk surface. The two $\cos\vartheta$ factors canceled each other.

Now let me consider an illuminated Lambertian sphere, illuminated from the viewing direction. Its irradiance (in W/m$^2$) will be proportional to $\cos\vartheta$:
\begin{align*} E=E_0\cos\vartheta. \end{align*}
It is related to its radiosity by its surface albedo $\alpha$:
\begin{align*} J=\alpha E=\alpha E_0\cos\vartheta. \end{align*}
It is still a Lambertian surface, therefore
\begin{align*} L_0=\frac{1}{\pi}J=\frac{1}{\pi}\alpha E_0\cos\vartheta. \end{align*}
Its radiance in the viewing direction is given by
\begin{align*} L=L_0\cos\vartheta=\frac{1}{\pi}\alpha E_0\cos^2\vartheta. \end{align*}
Multiplying by the surface element $d\Sigma$ gives
\begin{align*} L d\Sigma=L\frac{1}{\cos\vartheta}dA=\frac{1}{\pi}\alpha E_0\cos\vartheta dA. \end{align*}
If the projected surface of the sphere in the image plane is a disk with radius $R$, then a radial position on that disk is given by $r=R\sin\vartheta$, and consequently, $dr=R d\vartheta \dfrac{d}{d\vartheta}\sin\vartheta=R\cos\vartheta d\vartheta.$

The radiant intensity in the viewing direction is the integral of $L$ over the surface of this disk:
\begin{align*} I=\iint \frac{1}{\pi}\alpha E_0\cos\vartheta dA=\int_0^{2\pi}d\psi\int_0^R r dr\frac{1}{\pi}\alpha E_0\cos\vartheta=2\pi\int_0^{\pi/2}R^2\frac{1}{\pi}\alpha E_0\cos^2\vartheta\sin\vartheta d\theta=\frac{2}{3}R^2\alpha E_0. \end{align*}
An alternate way of calculating $I$ is by using $\cos\vartheta=\sqrt{1-r^2/R^2}$:
\begin{align*} I=2\pi\int_0^R r dr\frac{1}{\pi}\alpha E_0\sqrt{1-r^2/R^2}=\frac{2}{3}R^2\alpha E_0. \end{align*}
The average radiance is just the radiant intensity divided by the disk area $A=\pi R^2$:
\begin{align*} \langle L\rangle=\frac{I}{A}=\frac{2}{3\pi}\alpha E_0. \end{align*}

So this is how we get a factor of $\dfrac{1}{\pi}$ for a flat Lambertian surface and $\dfrac{2}{3\pi}$ for a Lambertian sphere illuminated from the viewing direction.