The question is suppose $X$ is a Poisson variable with a random mean, such that $X\sim \text{Poisson}(Y)$ and $Y \sim \exp(\lambda)$. Find the mean and variance of $X$.
By the law of total expectation, $E[X] = E[E[X|Y]]$ and I know that $E[X|Y]=E[X|\sigma(Y)]$. If I apply the definition of conditional expectation, I did not reach to any conclusion.
Is there any other way to find the expected value of $X$?
You're on the right track by applying the law of total expectation, since then $\mathbb{E}[X|Y] = Y$ (as $X|\{Y=j\} \sim \mathrm{Poi}(j) \implies \mathbb{E}[X|\{Y=j\}]=j$) and so
\begin{align*} \mathbb{E}[X] = \mathbb{E} \left [\mathbb{E}[X|Y] \right ] = \mathbb{E}[Y] = \frac{1}{\lambda}. \end{align*}
Next apply the law of total variance, to wit
\begin{align*} \mathbb{V}(X) &= \mathbb{V}(\mathbb{E}[X|Y]) + \mathbb{E}(\mathbb{V}(X|Y)) = \mathbb{V}(Y) + \mathbb{E}[Y] = \frac{\lambda+1}{\lambda^{2}}. \end{align*}
