Are all 4-manifolds $Pin^{\tilde{c}}$? It's known that all oriented 4-manifolds admit a $Spin^c$ structure, ie. a spin structure on $TX\oplus\mathcal{L}$ for some complex line bundle $\mathcal{L}$. 
A usual generalization of this structure to unorientable manifolds is to ask for a spin structure on $TX\oplus\mathcal{L}\oplus\mathcal{E}$ where $\mathcal{L}$ is again a complex line and now $\mathcal{E}$ is a real line bundle (which the whole bundle being oriented will force to be the orientation line). This is called a $Pin^c$ structure. In cohomology, it amounts to an integral lift of $w_2$.
Unfortunately, not all 4-manifolds admit a $pin^c$ structure, eg. $\mathbb{RP}^2 \times \mathbb{RP}^2$. This is easy to see by a computation of $w_2$.
There is another generalization I'll call a $Pin^{\tilde c}$ structure. In this version, $\mathcal{L}$ and $\mathcal{E}$ combine into a real 2-plane bundle. In cohomology it amounts to a twisted integral lift of $w_2$.
So, do all 4-manifolds admit a twisted integral lift of $w_2$?
Here are some edits in response to Qiaochu's comment. An integral lift of $w_2$ is an element of $H^2(X,\mathbb{Z})$ mapping to $w_2$ in $H^2(X,\mathbb{Z}/2)$. A twisted integral lift lives instead in $H^2(X,\mathbb{Z}^{w_1})$, with local coefficients in the orientation line.
 A: $\newcommand{\RP}{\mathbb{RP}}\newcommand{\Z}{\mathbb Z}$No, $\RP^2\times\RP^2$ isn't pin$\tilde c$. This came up while thinking about another MathOverflow
question, but I'll rewrite the argument here.
What you call a pin$\tilde c$-structure has been studied in physics, following
Metlitski and Freed-Hopkins, where it's
also called a pin$\tilde c+$-structure.  (This is because one can also ask about pin$\tilde
c-$-structures, which are twisted integral lifts of $w_2 + w_1^2$.) These structures are discussed in
a little more detail by Shiozaki-Shapourian-Gomi-Ryu in Appendix D.
Here's the argument for $\RP^2\times\RP^2$: let $x$ denote the generator of $H^1(-;\Z/2)$ of the first $\RP^2$ and
$y$ be that for the second $\RP^2$. Using the usual CW structure on $\RP^2$ and the product CW strucure on
$\RP^2\times\RP^2$, you can check that $H_2(\RP^2\times\RP^2;\Z)\cong\Z/2$ and the reduction mod 2 map
$H_2(\RP^2\times\RP^2;\Z)\to H_2(\RP^2\times\RP^2;\Z/2)$ sends the nonzero element of $H_2(\RP^2\times\RP^2;\Z)$ to
the Poincaré dual of $x+y$.
The Poincaré duality isomorphisms $H_2(\RP^2\times\RP^2;\Z)\cong H^2(\RP^2\times\RP^2;\Z_{w_1})$ and
$H_2(\RP^2\times\RP^2;\Z/2)\cong H^2(\RP^2\times\RP^2;\Z/2)$ are natural with respect to change-of-coefficients,
which means that the reduction mod 2 map $\rho_2\colon H^2(\RP^2\times\RP^2;\Z_{w_1})\to
H^2(\RP^2\times\RP^2;\Z/2)$ sends the nonzero element of $H^2(\RP^2\times\RP^2;\Z_{w_1})\cong\Z/2$ to $x+y$.
However, $w_2(\RP^2\times\RP^2) = x^2+xy+y^2$, so it's not in the image of $\rho_2$, and therefore
$\RP^2\times\RP^2$ has no pin$\tilde c+$-structure.
