You're looking for the score [tex]k[/tex] such that
[tex]\mathbb P(X\le k)=0.78[/tex]
First transform [tex]X[/tex] to the standard normal random variable.
[tex]\mathbb P(X\le k)=\mathbb P\left(\dfrac{X-1028}{92}\le\dfrac{k-1028}{92}\right)=\mathbb P(Z\le \hat k)=0.78[/tex]
Now 0.78 corresponds to a z-score of approximately [tex]\hat k=0.7722[/tex], which means
[tex]\dfrac{k-1028}{92}=\hat k\implies k\approx92(0.7722)+1028=1099.04\approx1099[/tex]
