You're looking for a value [tex]z[/tex] such that
[tex]\mathbb P(0<Z<z)=0.17\quad\text{or}\quad\mathbb P(-z<Z<0)=0.17[/tex]
Because the distribution is symmetric, the value of [tex]z[/tex] in either case will be the same.
Now, because the distribution is continuous, you have that
[tex]0.17=\mathbb P(0<Z<z)=\mathbb P(Z<z)-\mathbb P(Z<0)[/tex]
The mean for the standard normal distribution is [tex]0[/tex], and because the distribution is symmetric about its mean, it follows that [tex]\mathbb P(Z<0)=0.5[/tex].
[tex]0.17=\mathbb P(Z<z)-0.5\implies0.67=\mathbb P(Z<z)[/tex]
You can consult a [tex]z[/tex] score table to find the corresponding score for this probability. It turns out to be [tex]z\approx0.4399[/tex].
