Answer
$L_3 =L_0 +\dfrac{3mg}{k} $
Or
$ L_3 = 3 L_1-2L_0$
Work Step by Step
Let's assume that the stretching of the spring due to its own weight when suspended from the ceiling is negligible.
After attaching the mass $m$, the spring stretched a distance of $x=L_1-L_0\tag1$
We know that the system is at rest which means that the net force exerted on the mass $m$ is zero.
$$\sum F_y=F_{sp}-mg=ma_y=m(0)=0$$
Thus,
$$F_{sp}=mg$$
According to Hooke's law, $F_{sp}=-kx$
$$-kx=mg$$
The negative sign is due to the direction of stretching, so we can ignore it for now since we are dealing with its length magnitude.
Plugging from (1);
$$ k(L_1-L_0)=mg$$
$$ L_1-L_0 =\dfrac{mg}{k}\tag 2$$
$$ L_1 =L_0 +\dfrac{mg}{k}$$
Hence, when we attach a mass of $3m$;
$$ \boxed{L_3 =L_0 +\dfrac{3mg}{k}}$$
Solving (2) for $k$,
$$ k =\dfrac{mg}{L_1-L_0}$$
plugging into the boxed formula.
$$ L_3 =L_0 +\dfrac{3mg}{\dfrac{mg}{L_1-L_0}}=L_0 +\dfrac{3\color{red}{\bf\not} m\color{red}{\bf\not} g(L_1-L_0)}{\color{red}{\bf\not} m\color{red}{\bf\not} g} $$
$$ L_3 = L_0 + 3(L_1-L_0)= L_0 + 3 L_1-3L_0 $$
$$\boxed{ L_3 = 3 L_1-2L_0 }$$
