Step 1: Work with the magnetic moment instead of the pole strength alone.
For a bar magnet, the field at an equatorial point a distance $d$ from the centre is
\[
B_H = \frac{\mu_0}{4\pi}\frac{M}{\left(d^2+\left(\frac{l}{2}\right)^2\right)^{3/2}}, \qquad M = p\,l
\]
where $l$ is the magnet's length and $p$ its pole strength.
Step 2: Read off the geometry.
The two neutral points are $12\ \text{cm}$ apart and symmetric about the centre, so $d = 6\ \text{cm} = 0.06\ \text{m}$, and half the magnet's length is $l/2 = 8\ \text{cm} = 0.08\ \text{m}$.
\[
d^2+\left(\frac{l}{2}\right)^2 = 0.0036+0.0064 = 0.01\ \text{m}^2 \implies (0.01)^{3/2}=0.001\ \text{m}^3
\]
Step 3: Solve for the magnetic moment.
\[
M = \frac{4\pi B_H \times 0.001}{\mu_0} = \frac{B_H\times0.001}{10^{-7}} = 3.2\times10^{-5}\times10^4 = 0.32\ \text{Am}^2
\]
Step 4: Divide by the length to get the pole strength.
\[
p = \frac{M}{l} = \frac{0.32}{0.16} = 2\ \text{Am}
\]
\[
\boxed{2\ \text{Am}}
\]