The 85th percentile is the cutoff time [tex]t[/tex] such that
[tex]\mathbb P(X<t)=0.85[/tex]
In other words, the 85th percentile refers to the time needed to belong to the top 15% of the distribution; more generally, the [tex]n[/tex] percentile is the top [tex](100-n)\%[/tex] of the distribution.
Anyway, to find this value of [tex]t[/tex], transform [tex]X[/tex] to a random variable [tex]Z[/tex] with the standard normal distribution using
[tex]Z=\dfrac{X-\mu}\sigma[/tex]
where [tex]\mu[/tex] is the mean of [tex]X[/tex] and [tex]\sigma[/tex] is the standard deviation of [tex]X[/tex].
[tex]\mathbb P(X<t)=\mathbb P\left(\dfrac{X-5.9}{2.1}<\dfrac{t-5.9}{2.1}\right)=\mathbb P(Z<t^*)=0.85[/tex]
Here [tex]t^*[/tex] is used to denote the z-score corresponding to the cutoff time [tex]t[/tex]. Referring to a z-score table, you find that this occurs for [tex]t^*\approx1.036[/tex]. So,
[tex]\dfrac{t-5.9}{2.1}=t^*\implies t=5.9+2.1t^*\approx8.076[/tex]
