We know that pdf of normally distributed RV $f(x)=\frac{1}{\sigma \sqrt{2\pi }}e^{\frac{-(x-\mu )}{2\sigma ^{2}}}$
Obviously maximum probability will occur at the mean because mean is a point which is closest to all other data set points and therefore the probability of picking up points close to the mean is the most.
Therefore $x=\mu$, therefore our pdf = $f(x)=\frac{1}{\sigma \sqrt{2\pi }}e^{0}$
Therefore max probability = $\frac{1}{2.5\sqrt{2\pi }}$
Therefore for 1000 items = $\frac{1000}{2.5\sqrt{2\pi }}$=$\frac{400}{\sqrt{2\pi }}$