Now that you have all the Stiefel-Whitney classes written down, the hard part is over. To compute Stiefel-Whitney numbers, recall that these are, by definition, obtained in the following way.
Start with a partition of $4$, that is, a sum of a bunch of positive numbers which give $4$. Here are all five of the options: $1+1+1+1,\, 1+1+2,\, 1+3,\, 2+2,\,$ and $4$.
For each choice, form the corresponding product of Stiefel-Whitney class \begin{align*} 1+1+1+1 &\leftrightarrow w_1 \cup w_1 \cup w_1 \cup w_1 \\ 1+1+2 &\leftrightarrow w_1\cup w_1\cup w_2\\ 1+3 &\leftrightarrow w_1\cup w_3\\ 2+2 &\leftrightarrow w_2\cup w_2 \\ 4&\leftrightarrow w_4\end{align*}
The point of a partition is that all the cup products on the right land in $H^4(P^2\times P^2;\mathbb{Z}/2)$. Since every manifold has an orientation class mod $2$, we can pair the element on the right with the orientation class and get a number mod $2$ out. These numbers mod $2$ are the Stiefel-Whitney numbers.
By Poincare duality, the orientation class is the dual of the unique element in $H^4(P^2\times P^2,\mathbb{Z}/2)$, that is, it's the dual of $a^2 b^2$. Hence, computing all the Stiefel-Whitney numbers is the same as computing all the above cup products (using the relations $a^3 = b^3 = 0$), and then counting, mod $2$, the number of occurrences of $a^2 b^2$.
Doing this (while supressing the cup product sign) gives \begin{align*} (w_1)^4 &= (a+b)^4 & &= 0\\ (w_1)^2 w_2 &= (a+b)^2(a^2 + b^2 + ab) & &= 0 \\ w_1 w_3 &= (a+b)(ab^2 + a^2 b) & &= 0\\ (w_2)^2 &= (a^2+b^2+ab)^2 & &= a^2 b^2\\ w_4 &= a^2b^2 & &= a^2 b^2.\end{align*}
(Note that the computations are considerable eased by noting we're working mod $2$ so $(a+b)^2 = a^2 + b^2$.)
From this calculation, we see that three of the Stiefel-Whitney numbers are $0$ (mod $2$) while the other two are $1$ (mod $2$).
Suppose $M$ is embedded into the standard inner product space $\Bbb{R}^n$.
Since $TM \oplus NM \cong M \times \Bbb{R}^n$, by the Whitney sum formula, $w(TM) \cdot w(NM)=w(TM \oplus NM)=w(M \times \Bbb{R}^n)=1 \in H^0(M;\Bbb{Z}_2)$ where $w$ denotes the total Stiefel-Whitney class.
Specifically, $w_1(TM) + w_1(NM)=w_1(M \times \Bbb{R}^n)=0 \in H^1(M;\Bbb{Z}_2)$, which means $w_1(NM) = w_1(TM) \ne 0$.
Best Answer
The trivial vector bundle over any manifold has $w = 0$. The tangent bundle $TM$ of a nonorientable (smooth) manifold $M$ has $w(TM)\not = 0$, but that doesn't apply to arbitrary bundles over $M$.