Similar to your other question. Now, you're looking for the score [tex]k[/tex] such that
[tex]\mathbb P(X\le k)=0.99[/tex]
This time, 0.99 corresponds to a z-score of approximately [tex]\hat k=2.3264[/tex], which means
[tex]\dfrac{k-1028}{92}=\hat k\implies k\approx92(2.3264)+1028=1242.03\approx1242[/tex]
