Let $\pi \colon X \to \mathbb{P}^3$ be the first blow-up.

The normal bundle of $L_2$ in $\mathbb{P^3}$ is $$N_{L_2 / P^3}=\mathcal{O}_{P^1}(1) \oplus \mathcal{O}_{P^1}(1);$$ in fact $L_2$ moves into a family of dimension $4=h^0(N_{L_2/ P^3})$, namely the projective Grassmannian $\mathbb{G}(1,3)$.

The strict transform $L_2'$ of $L_2$, instead, moves into a family of smaller dimension, namely the strict transform of the family $\mathfrak{X}$ of lines intersecting $L_1$. Since $\mathfrak{X}$ is a divisor in $\mathbb{G}(1,3)$, we have $h^0(N_{L_2'/ X})=3$. In fact, one proves that
$$N_{L_2' / X}=\mathcal{O}_{P^1}(1) \oplus \mathcal{O}_{P^1},$$
see the edit at the endo of the post.

It follows that the exceptional divisor of the blow-up of $X$ along $L_2'$ is the projective bundle $$\mathbb{P}(\mathcal{O}_{P^1}(1) \oplus \mathcal{O}_{P^1}),$$
which is in turn isomorphic to the Hirzebruch surface $\mathbb{F}_1$.

**EDIT.** Let me prove in full details that $N_{L_2' / X}=\mathcal{O}_{P^1}(1) \oplus \mathcal{O}_{P^1}.$

Let us consider the family $\mathcal{S}$, given by the strict transforms in $X$ of the planes containing $L_2$, and let $S$ be a general element of $\mathcal{S}$. Then $S$ is isomorphic to $\mathbb{F}_1$, since it is just a plane blown-up at the point $L_1 \cap L_2$. Moreover $L_2' \subset S$ is a fibre of the ruling. Finally, since two elements of $\mathcal{S}$ intersect in a reducible curve made by a fibre and the $(-1)$-curve, it follows $$N_{S/X}=\mathcal{O}_S(S) \cong \mathcal{O}_{\mathbb{F}_1}(C_0+f),$$

where $C_0$ is the $(-1)$-curve and $f$ is a fibre of the ruling (of course, these $\mathbb{F}_1$ have nothing to do with the exceptional divisor of the second blow-up...)

Now, by using the normal bundle sequence associated with $L_2' \subset S \subset X$ we obtain
$$0 \to N_{L_2' / S} \to N_{L_2' / X} \to N_{S / X} \otimes \mathcal{O}_{L_2'} \to 0,$$
that is
$$ 0 \to \mathcal{O}_{P^1} \to N_{L_2' / X} \to \mathcal{O}_{P^1}(1) \to 0$$
(the last line bundle on the right comes from the fact that $(C_0+f)f=1$).

Since $\operatorname{Ext}^1(\mathcal{O}_{P^1}(1), \mathcal{O}_{P^1})=H^1(\mathcal{O}_{P^1}(-1))=0,$ the last exact sequence splits and the claim follows.