We want to find [tex]x[/tex] such that
[tex]\mathrm{Pr}(X \le x) = 0.81[/tex]
where [tex]X[/tex] is the random variable for test scores, and [tex]X\sim\mathrm{Normal}(63,9.3^2)[/tex].
Transform [tex]X[/tex] to [tex]Z\sim\mathrm{Normal}(0,1)[/tex] using the relation
[tex]X = \mu + \sigma Z \implies Z = \drac{X-63}{9.3}[/tex]
so that
[tex]\mathrm{Pr}(X \le x) = \mathrm{Pr}\left(\dfrac{X-63}{9.3} \le \dfrac{x-63}{9.3}\right) = \mathrm{Pr}\left(Z \le \dfrac{x-63}{9.3}\right) = 0.81[/tex]
Applying the inverse CDF of [tex]Z[/tex] (denoted by [tex]\Phi^{-1}[/tex]), we have
[tex]\dfrac{x-63}{9.3} = \Phi^{-1}(0.81) \approx 0.8779[/tex]
Solve for [tex]x[/tex].
[tex]\dfrac{x-63}{9.3} \approx 0.8779[/tex]
[tex]x-63 \approx 8.1644[/tex]
[tex]x \approx 76.1644[/tex]
Then the cutoff test score for the 81% percentile is [tex]P_{81} \approx \boxed{76}[/tex].
