$w_3(G)$ of a principal $G$ bundle over some 4-manifold $X_4$ I am puzzled by the possible non-zero $w_3(G)$ of a principal $SO(3)$-$G$ bundle over the 4-manifold $X_4$, and how can $w_3(G)$ be related to the lower dimensional $w_2(G)$. Here $w_i$ is the Stiefel Whitney classes over the principal $SO(3)$ bundle.
May we consider a few cases of examples, say $X_4=S^4,\mathbb{P}^4(\mathbb{R}),\mathbb{P}^2(\mathbb{C}),$ connected sum of them, etc. 

Can we construct some examples of non-zero $w_3(G)$ of a principal $SO(3)$-$G$ bundle over some compact 4-manifold $X_4$? How explicitly can they be related to the lower dimensional $w_2(G)$ and other characteristic class constraint?

 A: There are two relevant relations in the cohomology of $BSO(2n+1)$: first, the Euler class is 2-torsion, and in fact $e = \beta w_{2n}$; and second, the Euler class in the cohomology of $BSO(k)$ reduces to $w_k$ mod 2. Therefore, for a rank $2n + 1$ vector bundle $E$, $w_{2n+1}(E) = \beta w_{2n}(E) \mod 2$. 
This operation, taking the integral Bockstein $H^{k}(X;\Bbb Z/2) \to H^{k+1}(X; \Bbb Z)$, and composing with reduction mod 2 $H^{k+1}(X;\Bbb Z) \to H^{k+1}(X;\Bbb Z/2)$ is actually the first Steenrod square operation, written $\text{Sq}^1: H^*(X;\Bbb Z/2) \to H^{*+1}(X;\Bbb Z/2)$. 
Now the Dold-Whitney theorem says that on a 4-complex $X$, $SO(3)$-bundles are classified up to isomorphism by $(w_2, p_1) \in H^2(X;\Bbb Z/2) \times H^4(X;\Bbb Z)$, where these satisfy the constraint $w_2^2 = p_1 \pmod 4$ - note that here we are using the Pontryagin square to square a $\Bbb Z/2$ class into a $\Bbb Z/4$-class. In addition, there is always an $SO(3)$-bundle with a given pair $(w_2, p_1)$ as long as it satisfies this constraint.  
In any case, on an oriented closed manifold, $H^4(X;\Bbb Z) = \Bbb Z$, so there are $\Bbb Z$-many $SO(3)$-bundles with fixed $w_2$. 

Knowing this relationship and the Dold-Whitney classification, your question reduces to finding oriented examples where $\text{Sq}^1: H^2(X;\Bbb Z/2) \to H^3(X;\Bbb Z/2)$ is nonzero. Certainly this is false for $S^4, \Bbb{CP}^2$, since those have no third cohomology. Unfortunately, $\text{Sq}^1: H^2(\Bbb{RP}^4;\Bbb Z/2) \to H^{3}(\Bbb{RP}^4;\Bbb Z/2)$ is zero, even though both of these groups are nonzero. (Easiest way to see this is probably a by hand chain level calculation.)
(It is worth knowing that $e(E) = \beta w_2(E) \in H^3(X;\Bbb Z)$ is precisely the obstruction to writing $E \cong \Bbb R \oplus \lambda$, where $\Bbb R$ is the trivial line bundle and $\lambda$ an oriented plane bundle, and $2e(E) = 0$ is the obstruction to writing $E \cong \xi \oplus \lambda$ where now $\xi$ is a real line bundle and $\lambda$ is a 2-plane bundle; that is, it is always possible to split in this way.) 
As an example of an oriented manifold where this is true, try $S^1 \times \Bbb{RP}^3$. If $x$ denotes the generator of $H^1(S^1;\Bbb Z/2)$ and $y$ the generator of $H^1(\Bbb{RP}^3;\Bbb Z/2)$, then $\text{Sq}^1(xy) = xy^2$, which is nonzero; even better, the cup-square of $xy$ is zero (since $x^2 = 0$). So pick the bundle with $w_2 = xy$ and $p_1 = 0$; this has nonzero $\beta w_2$. There is probably an explicit description of this bundle but I haven't expended any effort to think of it.
I suspect the Enriques surface provides another example.
