Answer
68, which seems to be fine.
Work Step by Step
$\alpha=1-0.9=0.1.$ Using the table: $z_{\alpha/2}=z_{0.05}=1.645.$
Hence the sample size:$\left (\frac{z_{\alpha/2}\cdot \sigma}{E}\right)^2=\left (\frac{1.645\cdot 15}{3}\right)=68$, which seems to be fine.
