Both distributions are symmetric about their means, so it’s natural to pick the means to be the same.
So without loss of generality, we’ll assume both distribution have mean 0.
The question then becomes how to choose the scale parameters.
You could just set the two scale parameters to be the same, but that’s similar to the Greek letter fallacy, assuming two parameters have the same meaning just because they have the same symbol.
Because the two distributions have different tail weights, their scale parameters serve different functions.
One way to replace a Laplace distribution with a normal would be to pick the scale parameter of the normal so that both two quantiles match.
For example, you might want both distributions to have have 95% of their probability mass in the same interval.
I’ve written before about how to solve for scale parameters given two quantiles.
We find two quantiles of the Laplace distribution, then use the method in that post to find the corresponding normal distribution scale (standard deviation).
The Laplace distribution with scale s has densityf(x) = exp(-|x|/s)/2s.
If we want to solve for the quantile x such that Prob(X > x) = p, we havex = –s log(2 – 2p).
Using the formula derived in the previously mentioned post,σ = 2x / Φ-1(x)where Φ is the cumulative distribution function of the standard normal.
Related postsAdding Gaussian or Laplacian noise for privacyGenerating Laplace random samplesData privacy consulting The normal distribution is the canonical example of a thin-tailed distribution, while exponential tails are conventionally the boundary between thick and thin.
“Thick tailed” and “thin tailed” are often taken to mean thicker than exponential and thinner that exponential respectively.
 You could use a Gaussian mechanism rather than a Laplace mechanism for similar reasons, but this makes the differential privacy theory more complicated.
Rather than working with ε-differential privacy you have to work with (ε, δ)-differential privacy.
The latter is messier and harder to interpret.