Step 1: Recall formula for interference maxima.
For \(n\) slits, the intensity of the primary maximum is proportional to \(n^2\), i.e.,
\[
I_n \propto n^2
\]
Step 2: Take ratio for two and three slits.
\[
\frac{I_2}{I_3} = \frac{2^2}{3^2} = \frac{4}{9}
\]
Step 3: Conclusion.
Hence, \(I_2 = \dfrac{4}{9} I_3\).

At a particular temperature T, Planck's energy density of black body radiation in terms of frequency is \(\rho_T(\nu) = 8 \times 10^{-18} \text{ J/m}^3 \text{ Hz}^{-1}\) at \(\nu = 3 \times 10^{14}\) Hz. Then Planck's energy density \(\rho_T(\lambda)\) at the corresponding wavelength (\(\lambda\)) has the value \rule{1cm}{0.15mm} \(\times 10^2 \text{ J/m}^4\). (in integer)
[Speed of light \(c = 3 \times 10^8\) m/s]
(Note: The unit for \(\rho_T(\nu)\) in the original problem was given as J/m³, which is dimensionally incorrect for a spectral density. The correct unit J/(m³·Hz) or J·s/m³ is used here for the solution.)