$CP^2$ does not even immerse in R^6.
Proof: If such an immersion exists, then the normal Euler class has the property, that its square is the normal Pontrjagin class, and that is -3 times the signature (when evaluated on the fundamental homology class). But in $H^2(CP^2)$ there is no such a class $x$, for which $x^2$ evaluated on the fundamental class is -3. QED.
Moreover the following theorem is true (It is essentially due to Hughs)
In the 4-dimensional oriented cobordism group $\Omega_4 \approx Z$ precisely the even elements contain a manifold that admits an immersion into $R^6.$
The conditions (the manifold must be spin and have zero signature) are clearly necessary: An embedded manifold in a Euclidean space has zero normal Euler class. Hence in the present case both $p_1$ and $w_2$ are zero. The opposite is non-trivial, it is the containt of Ruberman's paper mentioned above.