Given data, $l=0.928\times {10}^{-2 }\mathrm{m}, A=1\times {10}^{-6} {\mathrm{m}}^2,$$n_i=2.5\times {10}^{19} {\mathrm{m}}^{–3}, {\mathrm{\mu }}_h=0.15 {\mathrm{m}}^2 {\left(\mathrm{V} \mathrm{s}\right)}^{-1}$ and ${\mathrm{\mu }}_e=0.35 {\mathrm{m}}^2 {\left(\mathrm{V} \mathrm{s}\right)}^{-1}$,

First, recall the relation between mobility and conductivity, 

$\sigma =\mu ne=\left({\mu }_h+{\mu }_e\right)n_{i}e$

$\Rightarrow \sigma =\left(0.15+0.35\right)\times 2.5\times {10}^{19}\times 1.6\times {10}^{-19}=2 \mathrm{mho} {\mathrm{m}}^{-1}$

We know that conductivity is the reciprocal of resistivity, 

$\Rightarrow \rho =\frac{1}{\sigma }=\frac{1}{2} \mathrm{\Omega } \mathrm{m}=50 \mathrm{\Omega } \mathrm{cm}$.