Solution :
Intensity of em wave $I= U_{av}c$
$U_{av}= \large\frac{1}{2} \frac{B_0^2}{M_0^2}$
$U_{av}$ (electric field) $= \large\frac{1}{2} \in_0$$ (c B_0)^2$
$\qquad= \large\frac{1}{2} $$\in_0 \large\frac{1}{\mu_0 \in _0}$$B_0^2$
$\qquad= \large\frac{1}{2} \frac{B_0^2}{\mu_0}$
$\qquad= U_{ar}$ (magnetic field)
This proves that in $em$ wave is divided equally between electric field vector and magnetic field vector .
Hence the ratio of contribution is $1:1$