Solution :
given $B= 12 \times 10^{-8} \sin (1.20 \times 10^7 z - 3.6 \times 10^{15})T$
$B= B_0 \sin (kz-wt)$
Comparing both the equation we get ,
$B_0 = 12 \times 10^{-8}T$
$I_{av}= \large\frac{1}{2} \frac{B_0^2}{\mu_0}$$c$
$\qquad = \large\frac{1}{2} \times \frac{12 \times 10^{-8}}{4 \pi \times 10^{-7}}$
$\qquad= 1.71\;w/m^2$