Answer
$N\geq e^{100}$
Work Step by Step
$\int_{N}^{\infty}\frac{1}{x(lnx)^{2}}dx\leq 0.01$
$[-\frac{1}{lnx}]_{N}^{\infty}\leq 0.01$
$-\frac{1}{ln\infty}-(-\frac{1}{lnN})\leq 0.01$
$\frac{1}{lnN}\leq \frac{1}{100}$
$lnN\geq 100$
Hence, $N\geq e^{100}$