A cube of unit volume contains $35\times {10}^7$ photons of frequency ${10}^{15}\text{Hz}$. If the energy of all the photons is viewed as the average energy being contained in the electromagnetic waves within the same volume, then the amplitude of the magnetic field is $\alpha \times {10}^{-9}\text{T}$. Taking permeability of free space ${\mu }_0=4\pi \times {10}^{-7}\text{Tm/A}$, Planck’s constant $h=6\times {10}^{-34}\text{Js}$ and $\pi =\frac{22}{7}$, the value of $\alpha$ is ______.

To find the value of $\alpha$, we can equate the total energy of the photons in the cube to the average electromagnetic energy contained in the wave within the same volume.

Step 1: Calculate the total energy of the photons.
The energy of a single photon of frequency $v$ is given by Planck's formula:
$E=hv$
Given:
Number of photons, $N=35\times {10}^7$
Frequency, $v={10}^{15}\text{Hz}$
Planck's constant, $h=6\times {10}^{-34}\text{Js}$
The total energy of all $N$ photons is:
$E_{\text{total}}=N·hv$
Substituting the given values:
$E_{\text{total}}=(35\times {10}^7)\times (6\times {10}^{-34})\times {10}^{15}$
$E_{\text{total}}=210\times {10}^{-12}\text{J}$

Step 2: Relate total energy to the average energy density of the electromagnetic wave.
The total volume of the cube is $V=1{\text{m}}^3$ (unit volume).
The average energy density $u$ of an electromagnetic wave is:
$u=\frac{E_{\text{total}}}{V}=210\times {10}^{-12}{\text{J/m}}^3$

The average energy density of an electromagnetic wave in free space in terms of the magnetic field amplitude $B_0$ is given by:
$u=\frac{B_0^2}{2{\mu }_0}$

Step 3: Solve for the magnetic field amplitude $B_0$.
Equating the two expressions for the average energy density:
$\frac{B_0^2}{2{\mu }_0}=210\times {10}^{-12}$
$B_0^2=2{\mu }_0\times 210\times {10}^{-12}$
Given the permeability of free space ${\mu }_0=4\pi \times {10}^{-7}\text{Tm/A}$ and $\pi =\frac{22}{7}$:
$B_0^2=2\times (4\times \frac{22}{7}\times {10}^{-7})\times 210\times {10}^{-12}$
$B_0^2=8\times \frac{22}{7}\times 210\times {10}^{-19}$
Since $\frac{210}{7}=30$:
$B_0^2=8\times 22\times 30\times {10}^{-19}$
$B_0^2=5280\times {10}^{-19}$
$B_0^2=528\times {10}^{-18}{\text{T}}^2$

Taking the square root on both sides:
$B_0=\sqrt{528}\times {10}^{-9}\text{T}$
$B_0\approx 22.98\times {10}^{-9}\text{T}$

Comparing this with the given format $B_0=\alpha \times {10}^{-9}\text{T}$, we find:
$\alpha =22.98$