Step 1:

Let $X$ be the random variable denoting the height of a student .

$X\sim N(64.5,(3.3)^2)$

Step 2:

To find the height below which 99% of the students lie.

Let $X_1$ be the height.

$P(X < X_1)=0.99$

Step 3:

Let $Z$ be the standard normal variate.

$Z=\large\frac{X-\mu}{\sigma}$

$\;\;\;=\large\frac{X-64.5}{3.3}$

Step 4:

When $X=X_1$

$Z=\large\frac{X_1-64.5}{3.3}$

$P(Z < Z_1)=0.99$

$\Rightarrow P(Z< Z_1)=0.5+0.49$

$\Rightarrow P(0< Z < Z_1)=0.49$

$Z=2.33$(from the table)

Step 5:

$Z=\large\frac{X_1-\mu}{\sigma}$

$\mu =64.5$

$2.33=\large\frac{X_1-64.5}{3.3}$

$X_1-64.5=(2.33)(3.3)$

$X_1=64.5+(2.33)(3.3)$

$\;\;\;\;=64.5+7.7$

$\;\;\;\;=72.2$inches