Could it have been:
$\large l_p = 1.61619926^{-35}m \small \quad \text{ (Planck Length)}$
At any rate, to create a black hole, you simply need enough energy density in a single area that its escape velocity (the speed at which the sums of $E_k$ and $E_p$ are $0$) is larger than the speed of light. As you should know, the photon has no mass. However, its momentum and energy are contributed to the mass of black holes curving spacetime in such a way that the escape velocity of light is not high enough to escape it. The energy of a photon, in accordance with its frequency is:
$$E=hf=pc$$
E=Energy
h=Planck's Constant
f=Frequency
p=Momentum
c = the speed of light constant
The Schwarzschild radius is the radius at which the mass of something would cause its escape velocity to be equal to the speed of light, $c$. The equation is
$$R = 2GM/c^2$$
Since the photon has no mass, we can rearrange Einsteins mass/energy equivaleny equation in the sense of mass as follows:
$$
E^2=(mc^2)^2+(pc)^2\\
E^2=m^2c^4+p^2c^2\\
\frac{E^2}{c^4+p^2c^2}=m^2\\
m=\sqrt\frac{E^2}{c^4+p^2c^2}\\
m=\frac{E}{c^2+pc}
$$
I calculate that the wavelength would be:
$$
R=\frac{2G{\frac {E}{c^2+pc}}}{c^2} \\
R=\frac{2GEc^2}{c^2+pc} \\
R=\frac{2GEc}{p} \\
R=\frac{2Ghfc}{p} \\
f=\frac{Rp}{2Ghc} \\
\lambda = \frac{1}{f} = 2\cdot G\cdot h \cdot c \cdot \frac{1}{Rp}
$$
Now this is just me tinkering around with some equations. I've probably done some illegal operations here, but in the interest of my amusement, that's the answer I came up with.
Thanks to @Hypnosifl for recommending the far simpler (possibly illegal) equation:
$$R=\frac{2Ghf}{c^4}$$
which doesn't require us to know momentum.
