From the formula of current gain, $\mathrm{\beta }={\left(\frac{{\mathrm{\Delta i}}_{\mathrm{C}}}{{\mathrm{\Delta i}}_{\mathrm{B}}}\right)}_{{\mathrm{V}}_{\mathrm{CE}}=10}$

${\mathrm{\Delta i}}_{\mathrm{C}}=\left(4.5-3\right) \mathrm{mA}$

$\Rightarrow {\mathrm{\Delta i}}_{\mathrm{C}}=1.5 \mathrm{mA}$

$\Rightarrow {\mathrm{\Delta i}}_{\mathrm{C}}=1.5\times {10}^{-3} \mathrm{A}$

${\mathrm{\Delta i}}_{\mathrm{B}}=\left(30-20\right) \mathrm{\mu A}$

$\Rightarrow {\mathrm{\Delta i}}_{\mathrm{B}}=10 \mathrm{\mu A}$

$\Rightarrow {\mathrm{\Delta i}}_{\mathrm{B}}=10\times {10}^{-6} \mathrm{A}$

$\Rightarrow {\mathrm{\Delta i}}_{\mathrm{B}}={10}^{-5} \mathrm{A}$

$\Rightarrow \mathrm{\beta }={\left(\frac{{\mathrm{\Delta i}}_{\mathrm{C}}}{{\mathrm{\Delta i}}_{\mathrm{B}}}\right)}_{{\mathrm{V}}_{\mathrm{CE}}=10}=\left(\frac{1.5\times {10}^{-3}}{{10}^{-5}}\right)$

$\Rightarrow \mathrm{\beta }=\left(1.5\times {10}^{-3}\times {10}^5\right)$

$\Rightarrow \mathrm{\beta }=150$