Step 1:

Let $X$ denote the amount of cosmic radiation and a person is exposed to $X \sim N(4.35,(0.59)^2))$

To find $P(X > 5.20)$

Step 2:

Let $Z$ be the standard normal variate,

$Z=\large\frac{X-\mu}{\sigma}$

When $X=5.20$

$Z=\large\frac{5.20-4.35}{0.59}$

$\;\;\;=\large\frac{0.85}{0.59}$

$\;\;\;=1.44$

Step 3:

$\therefore P(X > 5.20)=P(Z > 1.44)$

$\qquad\qquad\qquad=P(0 < Z < \infty)-P(0 < Z < 1.44)$

$\qquad\qquad\qquad=0.5-0.4251$

$\qquad\qquad\qquad=0.0749$