What would be the mse (mean squared error) of my scaled dataset on the original scale?
TLDR: You need to multiply your scaled mse by $(max - min)^2$.

Let's give our values some names: Let $y_n$ denote the dependant variable, $m$ be the number of observations
\begin{align*}
&amp; min := \min\limits_{n} y_n, \\
&amp; max := \max\limits_{n} y_n
\end{align*}
and let $p_n'$ be the (scaled) predicted value for $y_n$. The scaled value $y_n'$ for $y_n$ is given by
$$ y_n' = \frac{y - min}{max - min}. $$
So scaled mse that you calculated is given by
$$ 0.02 = mse_{\text{scaled}} = \frac1m \sum\limits_n |y_n' - p_n'|^2 = \frac1m \sum\limits_n |\frac{y - min}{max - min} - p_n'|. $$
In order to get the true mse, you want to rescale $y_n'$ to $y_n$ and $p_n'$ to $p_n$ using
$$ p_n' =  \frac{p_n - min}{max - min}.$$
This gives us
\begin{align*}
mse_{\text{scaled}}
&amp; = \frac1m \sum\limits_{n} |\frac{y_n - min}{max - min} - \frac{p_n - min}{max - min}|^2 \\
&amp; = \frac1m \sum\limits_{n} |\frac{y_n}{max - min} - \frac{min}{max - min} - \frac{p_n}{max - min} + \frac{min}{max - min}|^2 \\
&amp; = \frac1m \sum\limits_{n} |\frac{y_n}{max - min} - \frac{p_n}{max - min}|^2 \\
&amp; = \frac{1}{(max - min)^2} \frac1m \sum\limits_{n} |y_n - p_n|^2.
\end{align*}
Rearraging this gives us
$$ mse_{\text{original}} = \frac1m \sum\limits_{n} |y_n - p_n|^2 = (max - min)^2 mse_{\text{scaled}}. $$

However, in general, it is better to first rescale your values and then take the mse because you lose interpretability and, as you see, it is more difficult to rescale the mse than to rescale the predicted values.