Acid value of linseed oil
Unfortunately, the proportion you are using is not correct. Acid value $\mathrm{AN}$ determination procedure is not much different from an ordinary titration with indicator with the only difference that the mass of titrant is normalized to the mass of oil $m_\mathrm{oil}$:

$$\mathrm{AN} = \frac{T\cdot ΔV_\mathrm{eq}}{m_\mathrm{oil}}$$

where $ΔV_\mathrm{eq}$ is equivalent volume (ideally you have to use a difference with an external standard ("spiking solution"), but you probably didn't do that); $T$ is a so-called titer, or basically mass concentration of the titrant:

$$T = CM$$

so that

$$
\begin{align}
\mathrm{AN} &amp;= \frac{C(\ce{KOH})\cdot M(\ce{KOH})\cdot ΔV_\mathrm{eq}}{m_\mathrm{oil}}\\
  &amp;= \frac{\pu{9.5e-3 mol L-1}\cdot\pu{56.106 g mol-1}\cdot\pu{7.4e-3 L}}{\pu{1.015 g}}\\
  &amp;= \pu{3.89 mg g-1}
\end{align}
$$

Also, a couple of side notes:


Always carry units across all steps and never omit them.  
I recommend not to do intermediate calculations if you don't have to. Instead, first find an algebraic equation and only then plug in all the numbers.
Mind significant figures. If the mass of oil is given with 4 significant figures, it's a good idea to use a molecular mass of the titrant with at least 4 sigfigs too.