Given an Lie Algebra $L$ (of finite dimension and over an algebraically closed field with zero characteristic) and an ideal $I$, is it truth that
$rad\left(\dfrac{L}{I}\right)= \pi(rad(L))$,
where $\pi$ is the projection? What happens if $L$ isn't finite dimensional? And if the field has positive characteristic?
A counter-example for the infinite dimensional case:
Let $V$ be a vector space with a basis $\{e_i\mid i\in \mathbb{Z}_+\}$ and $V_p=\langle e_i\mid i=1,\dots,p\rangle$. Consider $$\mathfrak{t}_\infty:= \{ x \in \mathfrak{gl}(V)\mid x(V_i)\subset V_i \} .$$ as a lie subalgebra of $\mathfrak{gl}(V)$.
Let $E_{ij}$ be the linear transformations on $V$ such that $E_{ij}(e_j)=e_i$ and $E_{ij}(e_p)=0$ if $p\neq j$. Then, we have the formula $$[E_{ij},E_{kl}]=\delta_j^kE_{il}-\delta_l^iE_{kj}.$$
Convince yourself that $$\mathfrak{h}:=\{x \in t_\infty \mid x(V) \subset \langle e_i\mid i>1\rangle \}$$ is an ideal of $\mathfrak{t}_\infty$.
So, we have that $$rad( \mathfrak{t}_\infty / \mathfrak{h})=$$
$$\langle E_{1j} \mid j>1\rangle / \mathfrak{h}.$$
Suppose, by absurd, that $$\pi ( rad(\mathfrak{t}_\infty) ) = $$
$$rad( \mathfrak{t}_\infty / \mathfrak{h}) =$$
$$\langle E_{1j}\mid j>1\rangle / \mathfrak{h}.$$ This implies that $$\langle E_{1j}\mid j>1\rangle + \mathfrak{h} = $$
$$rad(\mathfrak{t}_\infty) + \mathfrak{h}.$$
Therefore, $$\langle E_{1j}\mid j>1\rangle + \mathfrak{h} \ni E_{12}=$$
$$x + y \in rad(\mathfrak{t}_\infty) + \mathfrak{h}$$
for some $x \in rad(\mathfrak{t}_\infty)$ and $y \in \mathfrak{h}$. Then
$$rad(\mathfrak{t}_\infty) \ni $$
$$[E_{11},x]=$$
$$[E_{11},E_{12}]-[E_{11},y]=$$
$$E_{12}+0=E_{12}.$$
This, again by the formula given above, implies that
$$rad(\mathfrak{t}_\infty)$$
contains
$$t_\infty ':=\{x \in \mathfrak{t}_\infty\mid x(V_1)=0\textrm{ and } x(V_i)\subset V_{i-1}, i>1\}$$ that is a non-solvable ideal of $\mathfrak{t}_\infty$. A contradiction.
I am not sure about infinite dimensions and/or positive characteristic, but the answer is Yes in the finite-dimensional, zero characteristic case.
In your situation we have an exact sequence of Lie algebras $$ 0 \longrightarrow I \longrightarrow L \stackrel{\pi}{\longrightarrow} L/I \longrightarrow 0 $$ Let $r_L < L$ be the radical of $L$. Then $\pi(r_L)$ is a solvable ideal of $R/I$ and hence it is contained in its radical $r_{L/I}$, so that $$ \pi(r_L) < r_{L/I} $$
My original answer had an error, which Kevin Ventullo pointed out in a comment below. He also kindly fixed it in his second comment, which I include here for completeness, but the credit is due solely to him.
The Levi factor of $L$, which is semisimple and isomorphic to $L/r_L$, surjects onto $(L/I)/\pi(r_L)$, whence $(L/I)/\pi(r_L)$ has to be semisimple as well, whence we get the other inclusion $$ r_{L/I} < \pi(r_L) $$
