You need to know the distribution's mean and standard deviation.
[tex]\mathbb P(X>100)=0.20\implies\mathbb P(X\le100)=0.80[/tex]
Transforming to the standard normal distribution, and letting [tex]\sigma[/tex] denote the standard deviation,
[tex]\mathbb P(X\le100)=\mathbb P\left(\dfrac{X-86}\sigma\le\dfrac{100-86}\sigma\right)=0.80[/tex]
The z-score corresponding to this probability is approximately [tex]z=0.8416[/tex], so you have
[tex]\dfrac{100-86}\sigma=0.8416\implies\sigma=16.64[/tex]
