When transforming a normal distribution to a uniform distribution preserving both the mean and the standard deviation, the bounds of the uniform distribution must be the original mean plus or minus a constant scaling factor:
$$ a = \mu - \sqrt 3 \sigma $$ $$ b = \mu + \sqrt 3 \sigma $$
How is the $ \sqrt 3 $ found?
The density function $f(x)=\frac{1}{2\sqrt{3}\sigma}$ for $a\le x\le b$. Thus the variance $=\frac{1}{2\sqrt{3}\sigma}\int_a^b (x-\mu)^2dx=\sigma^2$.