Solution

Light incident normally on the metallic mirror reflects and interferes with the incident light, generating a standing electromagnetic wave. As shown in the previous problems, the electric field vanishes at the surface of an ideal metal—a node of the electric field ($E = 0$) is formed there.
The antinodes (maxima of electric field intensity) are located at distances of $\lambda/4$, $3\lambda/4$, $5\lambda/4$, etc., from the surface. The photographic emulsion is sensitive specifically to the electric component of the field, so its darkening occurs at the locations of these antinodes.

Wavelength:
The first darkening (first antinode) occurs at a distance $d_1 = 10^{-5}\text{ mm}$. Since $d_1 = \lambda/4$:
$$\frac{\lambda}{4} = 10^{-5}\text{ mm} \implies \lambda = 4 \cdot 10^{-5}\text{ mm}$$
(For reference: this is $4 \cdot 10^{-8}\text{ m} = 40\text{ nm}$, which corresponds to ultraviolet radiation).

Distance to the second layer:
The second layer of blackened emulsion forms at the location of the second antinode of the standing wave. The distance from the metal surface to the second antinode is:
$$d_2 = \frac{3\lambda}{4} = 3 \cdot d_1 = 3 \cdot 10^{-5}\text{ mm}$$