Answer
$n=3394$
Work Step by Step
$level~of~confiance=(1-α).100$%
$98$% $=(1-α).100$%
$0.98=1-α$
$α=0.02$
$z_{\frac{α}{2}}=z_{0.01}$
If the area of the standard normal curve to the right of $z_{0.01}$ is 0.01, then the area of the standard normal curve to the left of $z_{0.01}$ is $1−0.01=0.99$
According to Table V, the z-score which gives the closest value to 0.99 is 2.33.
Now, the sample size (no prior estimate of $p ̂$) :
$E=0.02$ (within 2 percentage points)
$n=0.25(\frac{z_{\frac{α}{2}}}{E})^2$
$n=0.25(\frac{2.33}{0.02})^2$
$n=3393.06$
Round up:
$n=3394$
