WinBUGS uses “precision” as a parameter in specifying a Normal distribution, instead of variance.

Precision = $\frac{1}{\text{Variance}}$

`dnorm(0, 0.0001)` is the same as a Normal distribution with mean
0 and variance $\frac{1}{0.0001} = 100^2$, or $\sim \mathcal{N}(0,100^2)$

# Edit

In response to the question, I assume your model looks something like this...

    model { 
       for (i in 1:n) { 
          y[i] ~ dnorm(mu[i], tau) 
          mu[i] <- a + b*x[i]
       }

       a ~ dnorm(0, 1.0E-6)
       b ~ dnorm(0, 1.0E-6)
    
       tau ~ dnorm(0,1.0E-6)
    
       sigma <- 1/sqrt(tau)
    }

So, you simulate a $\tau$ value as $\sim \mathcal{N}(0, \frac{1}{1.0E-6})$, then use that $\tau$ value to calculate $y[i]$, such that $y[i] \sim \mathcal{N}(0, \frac{1}{\mathcal{N}(0, \frac{1}{1.0E-6})})$

Using a mean-zero Gaussian is standard practice for most people simulating random positive-or-negative errors.  I do not know why you chose $1.0 \times 10^{-6}$ as your precision parameter. It is your model, not mine.