Conversion from a PPB value to µg/m3 of Isobutylene
This is much simpler than you think, even though the value you've obtained looks reasonable to me.
Part per billion ($\pu{1 ppb} = \pu{1e-9}$) on its own is meaningless.
Based on a context, it can refer to anything: mass fraction, mole fraction, particles etc.
The thing is, for gaseous mixtures $\pu{ppb}$ always refers to the volume fraction $\phi_i$:

$$\varphi_i = \frac{V_i}{V}$$

where $V_i$ is the volume of the $i$th gas (here, isobutylene) and $V$ is the total volume.
In order to convert volume fraction expressed in $\pu{ppb}$ to desired mass concentration $\rho_i$ expressed as

$$ρ_i = \frac{m_i}{V}$$

where $m_i$ is the mass of $i$th component in a mixture, you have to use density $d_i$:

$$ρ_i = \frac{m_i}{V} = \frac{m_i}{V_i}\varphi_i = d_i\varphi_i$$

To conform with your units of choice ($\pu{μg m-3}$) we need to juggle the units a bit as the density is usually given in $\pu{kg m-3}$, e.g. for gaseous isobutylene at NTP $d(\ce{C4H8}) = \pu{2.3959 kg m-3}$ (Source):

$$
\begin{align}
\pu{1 g} &amp;= \pu{1e6 μg} \\
d(\ce{C4H8}) &amp;= \pu{2.3959e6 μg m-3}
\end{align}
$$

Now, using this density value you can multiply your volume fraction in $\pu{ppb}$ or $\pu{ppm}$ and get an answer, e.g. for $\pu{400 ppm}$:

$$ρ_i = d(\ce{C4H8})\varphi (\ce{C4H8}) = \pu{2.3959e6 μg m-3}\cdot\pu{400e-6} = \pu{958 μg m-3}$$

Note that all notations involving "part per something" are deprecated and should be avoided.
It also wouldn't hurt if you ask manufacturer whether $\pu{1 ppb}$ the sensor reports is indeed $\pu{1e-9}$ and not $\pu{1e-12}$ (just in case) if you think the reported values are too high.