Conductivity of Wüstite
after revisiting the problem I realised where I went wrong. The main mistake with not being careful with units and not reading the qurstion and having incorrect values for hole mobility. 
$$\text{mass}_{\ce{Fe(III)}}=((0.24)(55.85))/((N_\mathrm{a})(10^3))$$
$$=2.225838592\times10^{-26}\: \mathrm{kg}$$
$$\rho_{\ce{Fe(III)}}=2.225838592\times10^{-26}/(4.37\times10^{-10})^3$$
$$=\mathrm{266.7\: kg\: m^{-3}}$$
$$\text{No. of holes}=\rho_{\ce{Fe(III)}}/1\:\text{AMU}_\ce{{Fe(III)}}$$
$$\Rightarrow \text{No. of holes}=266.7/(9.296\times10^{-26})$$
$$=2.868975904\times\pu{10^{27}\:holes/m^3}$$
Conductance for a p-type semiconductor is given as: $$\sigma=p(q)(\mu_{h})$$
$$\implies\sigma=(2.868975904\times10^{27})(1.6\times10^{-19})(3.1\times10^{-5})$$
$$\approx1.4\times10^{4}\:(\Omega.m)^{-1}$$