When asked to derive the distribution of a random variable it's customary to present the cumulative distribution function (cdf), commonly denoted $F_Y(x):=\mathbb{P}(Y\leq x)$, for r.v. $Y$. In your case, it is probably helpful to note that $\mathbb{P}(Y\leq x)=1-\mathbb{P}(Y> x)$. Now, the minimum of 3 variables is of course greater than $x$ exactly when (iff) all of them are greater than $x$. You then get that $\mathbb{P}(Y> x)=\mathbb{P}(X_1>x, X_2>x,X_3>x)=\mathbb{P}(X_1>x)\mathbb{P}(X_2>x)\mathbb{P}(X_3 >x)$, where the last step follows from independence of the $\{X_i\}$.
Putting things together, $\mathbb{P}(Y\leq x)=1-\mathbb{P}(X_1>x)\mathbb{P}(X_2>x)\mathbb{P}(X_3 >x)$. Note now that $\mathbb{P}(X_i >x)=e^{-\lambda_ix},\forall i$ and you can probably fill in the last details yourself, i.e. simplify and note that $Y$ is also exponentially distributed and find its parameter.
As pointed out by @Drew75 in the comments, one should keep in mind that the mean of an exponential random variable with parameter $\lambda$ is equal to $1/\lambda$.