Is it possible to express a vector using cross product with another vector?

$\underline{a}$ would have to be a vector perpendicular to $\underline{c}$ but not parallel to $\underline{b}.$

You would therefore have to find parameters $\lambda$ and $\mu$ so that, for example $$\underline {a}=\lambda\underline{b}+\mu\underline{b}\times\underline{c}$$ where $\mu\neq0.$

We then have $$\underline{a}\times\underline{b}=(\lambda\underline{b}+\mu\underline{b}\times\underline{c})\times\underline{b}$$ $$=0+\mu(\underline{c}(b\cdot b)-\underline{b}(b\cdot c))$$

But $b\cdot c=0$ since they are perpendicular, so $\mu$ is predetermined and has value $\frac{1}{(b\cdot b)}$ and $\lambda$ is arbitrary.

So while $\underline{a}$ is not unique, it can be found for specific non parallel vectors $\underline{b}$ and $\underline{c}$

As mentioned elsewhere, only the components of $\vec{a}$ perpendicular to $\vec{b}$ can be found.

Use $$ \vec{a}_\perp = \frac{\vec{b}\times\vec{c}}{\| \vec{b} \|^2} $$

Proof

Vector $\vec{a}$ is assumed to contain both perpendicular and parallel components to $\vec{b}$ such that $\vec{a} = \vec{a}_\parallel + \vec{a}_\perp$.

Now form $\vec{c} = \vec{a} \times \vec{b}$

$$ \vec{c} = (\vec{a}_\parallel + \vec{a}_\perp) \times \vec{b} $$

and use the formula above

$$ \vec{c} = \left(\vec{a}_\parallel + \frac{\vec{b}\times\vec{c}}{\| \vec{b} \|^2}\right) \times \vec{b} $$

Since $\vec{a}_\parallel$ is parallel to $\vec{b}$, it stands that $\vec{a}_\parallel \times \vec{b} =0 $

$$ \vec{c} = \frac{ (\vec{b}\times\vec{c})\times \vec{b} }{\vec{b} \cdot \vec{b}} = \frac{\vec{b} \times (\vec{c} \times \vec{b} )}{\vec{b} \cdot \vec{b}} = \frac{\vec{c} (\vec{b} \cdot \vec{b}) - \vec{b} (\vec{c}\cdot\vec{b}))}{\vec{b} \cdot \vec{b}} = \vec{c}$$

which is an application of the vector triple product. The above is true since $\vec{c}\cdot\vec{b}=0$ with them being perpendicular to each other.