The columns of the matrix that represents the transformation in the given basis express the weights of the image of each of the basis vectors when those images are expressed in terms of the basis. That's quite a mouthful, so let's unpack what it means.
For the first column, consider $L(\vec{u}_1)$ in terms of $B$, which can be read off from the defining formula by specifying $c_1 = 1$ and the other coefficients $c_2 = c_3 = 0$: $$ L(\vec{u}_1) = \vec{u}_1 + 2\vec{u}_2, $$ so the first column of $M_B$ is $$ \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}. $$
Try to calculate the other two columns yourself before revealing:
$$ M_B = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 0 & 1 \\ 0 & -2 & -1 \end{bmatrix}. $$
By the way, notice that there's an alternative way that's easier to write down for this same matrix, with the linear transformation defined the way it is. Each of the components in the image specify the rows of the matrix. For example, the second row is $\begin{bmatrix} 2 & 0 & 1 \end{bmatrix}$, which you can read off of the second component: $$ (2c_1 + c_3) \, \vec{u}_2. $$ You can use this to fill out the matrix $M_B$ by rows.
Comment: the specification of the basis $B$ in terms of the standard coordinates is a red herring! It's completely irrelevant to the task at hand. We only care about those coordinates if we need to convert between the given basis $B$ and the standard basis. Only then does it become relevant that, for instance, $$ \vec{u}_2 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, $$ which says that $\vec{u}_2 = \vec{e}_1 + \vec{e}_2$, where $$ \bigl\{ \vec{e}_1, \vec{e}_2, \vec{e}_3 \bigr\} $$ is the standard basis.
