 Solution

In a large periodic quantization volume $\mathcal{V}$, one wavevector state occupies $(2\pi {)}^3/\mathcal{V}$ in wavevector space. There are two transverse photon polarizations, and each oscillator contributes $\hslash ck/2$. A lower wavelength cutoff ${\lambda }_{\min }$ corresponds to $k_{\max }=2\pi /{\lambda }_{\min }$, so the energy density isFor ${\lambda }_{\min }=10^{-10}\mathrm{m}$,The factor two is polarization, not an extra independent count of positive and negative frequencies. The continuum wavevector integral already counts independent travelling-wave modes in the usual field quantization convention. The quartic cutoff dependence also shows why an assumed shortest wavelength is crucial.