There are a few ways to tackle this problem. Nevertheless, one important issue is that the magnitude of the observation is 2. That is, $|x|=2$, which does NOT distribute Gaussian. In fact, for a zero-mean Gaussian random variable $x$ with variance $\sigma^2$, the distribution of its magnitude $y=|x|$ is the half-normal distribution
$$p_Y(y)=\begin{cases}\displaystyle{\frac{\sqrt{2}}{\sqrt{\pi}\sigma}\exp\bigg(-\frac{y^2}{2\sigma^2}\bigg)}&y\geq0\\0&\text{otherwise.}\end{cases}
$$
The second issue is that the random variables are continuous, and then $\mathbb{P}(X=a)=0$ for any $a$, and also $\mathbb{P}(|X|=a)=0$. What we will do is the following first-order approximation:
$$
\mathbb{P}(X=a)=\mathbb{P}(X\leq a+\delta) - \mathbb{P}(X\leq a) \approx p_X(a)\cdot\delta
$$
which comes from the fact that the pdf is the derivative of the CDF, used for expressions such as $\mathbb{P}(X\leq a)$, and where we assume $\delta$ is small.
Hence, we now compute the probability that transmitted signal $S$ was signal 1, given that the observed magnitude is $Y=2$. Knowing that signals are transmitted with equal frequency,
$$
\begin{align}
\mathbb{P}(S=1|Y=2)&=\frac{\mathbb{P}(Y=2|S=1)\mathbb{P}(S=1)}{\mathbb{P}(Y=2)}\quad\text{(Bayes' rule)}\\&=\frac{\mathbb{P}(Y=2|S=1)\mathbb{P}(S=1)}{\mathbb{P}(Y=2|S=1)\mathbb{P}(S=1)+\mathbb{P}(Y=2|S=2)\mathbb{P}(S=2)}\quad\text{(total prob.)}\\
&=\frac{\mathbb{P}(Y=2|S=1)}{\mathbb{P}(Y=2|S=1)+\mathbb{P}(Y=2|S=2)}\quad\text{(equal frequency of signals)}\\
&\approx \frac{p_Y(2|S=1)}{p_Y(2|S=1)+p_Y(2|S=2)}\quad\text{(approx. and $\delta$ vanishes in ratio)}\\
&=\frac{\frac{\sqrt{2}}{\sqrt{\pi}2}\exp\Big(-\frac{2^2}{2\cdot 4}\Big)}{\frac{\sqrt{2}}{\sqrt{\pi}2}\exp\Big(-\frac{2^2}{2\cdot 4}\Big)+\frac{\sqrt{2}}{\sqrt{\pi}3}\exp\Big(-\frac{2^2}{2\cdot 9}\Big)}\quad\text{(half-normal distributions)}\\
&=0.531878531904...
\end{align}
$$
