A very fundamental question about average normalized power of noise
If a filter with transfer function $H(f)$ has a noise process as input, then
the output noise power $P_{\scriptstyle{\text{out}}}$ is given by
$$P_{\scriptstyle{\text{out}}} = 
\int_{-\infty}^\infty |H(f)|^2 S_x(f)\,\mathrm df
$$
where $S_x(f)$ is the power spectral density of the input noise process.
If the input process is modeled as a white noise process, then $S_x(f)$
is a constant, call it $K$ for now, and so we have that
$$P_{\scriptstyle{\text{out}}} = K
\int_{-\infty}^\infty |H(f)|^2\,\mathrm df
$$
For an ideal low-pass filter with cutoff frequency $B$ Hz,
$$ H_{\scriptstyle{\text{lowpass}}}(f)
= \begin{cases} 1, &amp;-B \leq f \leq B,\\
0, &amp; |f| &gt; B,\end{cases}$$
we have $P_{\scriptstyle{\text{out}}} = 2KB$. More generally,
for an arbitrary linear time-invariant system with maximum
passband gain equal to $1$, the noise-equivalent bandwidth of the system
is defined as
$$B_{\scriptstyle{\text{ne}}} = \frac{1}{2}\int_{-\infty}^\infty |H(f)|^2\,\mathrm df$$
and so we get $P_{\scriptstyle{\text{out}}} = 2KB_{\scriptstyle{\text{ne}}}$.
Put another way, the noise-equivalent bandwidth of an ideal LPF with
cutoff $B$ Hz is just $B$ Hz.

It is conventional in the engineering literature to write $N_0/2$ instead
of $K$ because then the output noise power expression becomes
$N_0B_{\scriptstyle{\text{ne}}}$. The constant of proportionality
$N_0$, measured in watts/Hz as Jason R pointed out, is convenient
for use in ordinary language when conversing with other engineers.
If we have a filter with noise
equivalent bandwidth $B_{\scriptstyle{\text{ne}}}$, and
the thermal noise power at the input of the filter  is $N_0$ watts/Hz,
then the output noise power is $N_0B_{\scriptstyle{\text{ne}}}$.
That's the way people talk during discussions. Nobody will say that an ideal
lowpass filter with cutoff $B$ Hz has "bandwidth" $2B$ even though
the transfer function has value $1$ for $f \in [-B,B]$ and thus
the support of $H_{\scriptstyle{\text{lowpass}}}(f)$ has length
$2B$.

Turning to your question, it is not a good idea to write $W(f) = N_0$
for $f \in [-B,B]$, but if the paper you are reading does insist on this notation,
then the output power is $N_0(2B)$. If the paper says that it is
$\frac{1}{4}N_0(2B)$, then it is mistaken. Another possibility is 
that there are other conditions that are mentioned in the paper that
make $\frac{1}{4}N_0(2B)$ the correct answer, but you
have not included these conditions in your question. For example,
if the passband gain of the ideal LPF is $\frac{1}{2}$, then
the expression $\frac{1}{4}N_0(2B)$ for the output power
is perfectly correct.