If the total average energy density of an electromagnetic wave...

The total average energy density $U$ of an electromagnetic wave is the sum of the average electric energy density ($u_E$) and the average magnetic energy density ($u_B$).

Since $u_E = u_B$, the total average energy density is $U = 2u_B$.

The average energy density of the magnetic field is given by $u_B = \dfrac{B_{rms}^2}{2\mu_0}$.

Since $B_{rms} = \dfrac{B_0}{\sqrt{2}}$, we have $u_B = \dfrac{B_0^2}{4\mu_0}$.

Therefore, the total average energy density is:
$U = 2 \times \left(\dfrac{B_0^2}{4\mu_0}\right) = \dfrac{B_0^2}{2\mu_0}$

Rearranging for the amplitude of the magnetic field $B_0$ gives:
$B_0^2 = 2\mu_0 U \implies B_0 = \sqrt{2\mu_0 U}$.

Answer: $\sqrt{2\mu_0 U}$