(b) Current gain = $\frac{small\ change\ in\ collector\ current\ \left( {\mathrm{\Delta}i}_{c} \right)}{small\ change\ in\ emitter\ current\ \left( {\mathrm{\Delta}i}_{c} \right)}$
0.9 = $\frac{{\mathrm{\Delta}i}_{c}}{5}$
$\Rightarrow {\mathrm{\Delta}i}_{c}$= 0.9 $\times$ 5 = 4.5 mA