The smallest GPA needed to fall in the top 20% is [tex]g[/tex], where
[tex]\mathbb P(G\ge g)=0.20[/tex]
where [tex]G[/tex] is a random variable denoting GPA scores. Transform [tex]G[/tex] to the standard normal random variable [tex]Z[/tex] using
[tex]Z=\dfrac{G-\mu_G}{\sigma_G}\iff G=\mu_G+\sigma_GZ[/tex]
where [tex]\mu_G[/tex] and [tex]\sigma_G[/tex] are the mean and standard deviation of [tex]G[/tex], respectively. Now
[tex]\mathbb P(G\ge g)=\mathbb P\left(Z\ge\dfrac{g-2.29}{1.29}\right)=0.20[/tex]
Using the inverse CDF for the standard normal distribution, we find that the corresponding critical value is about 0.8416, which means
[tex]\implies\dfrac{g-2.29}{1.29}\approx0.8416\implies g\approx3.3757[/tex]
