Since \(L(t)\) is a constant it can be brought outside the integrand and the integral can be separated into a sum of integrals. Thus we're left with
\(X(t)=L(t)\large(\int_0^T\mu dt+\int_0^T\sigma dW_t\right)=L(t)\large(\mu T+\sigma\int_0^TdW_t\right)\)
Then, taking the expected value of this gives
\(E[X(t)]=L(t)\large(E[\mu T]+\sigma E\left[\int_0^TdW_t\right]\right)\)
Since \(L(t),\;\mu,\;T,\;\sigma\) are all constants they can be brought outside the expected value and all we're left with is the expected value of the integral of \(dW_t\), which is just 0 because the Ito integral is a Martingale. Thus we're left with
\(E[X(T)]=L(t)\mu T\)
We then need to calculate the second (non-central) moment to get the variance. I'm going to skip a couple of algebraic details.
\(E[X(t)^2]=L(t)^2\large(E\left[\large(\int_0^T\mu dt\right)^2\right]+E\left[\large(\int_0^T\sigma dW_t\right)^2\right]\right)\\E[X(t)^2]=L(t)^2\large(\mu^2T^2+\sigma^2\int_0^Tdt\right)\\E[X(t)^2]=L(t)^2(\mu^2T^2+\sigma^2T)\)
The first integral is evaluated going from line 1 to 2 above because it is deterministic just like in the first moment, and the second integral comes from Ito Isometry (since a constant is a square integrable function). Then the deterministic integral is just evaluated.
So we already know what the expected value is, and the variance is simply \(L(T)^2(\mu^2T^2+\sigma^2T)-(L(T)\mu T)^2=L(T)^2\sigma^2T\)
